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https://archive.org/details/treatiseonshadesOO0davi_1 


TREATISE 


ON 


SHADES AND SHADOWS, 


LINEAR PERSPECTIVE. 


BY CHARLES DAVIES, 


PROFESSOR OF MATHEMATICS IN THE MILITARY ACADEMY, 
AND 
AUTHOR OF THE COMMON SCHOOL ARITHMETIC, DESCRIPTIVE GEOMETRY, 
AND ELEMENTS OF SURVEYING. 


SECOND EDITION. 


PUBLISHED BY 


WILEY & PUTNAM, COLLINS, KEESE & CO., New-Yorx,—STATIONERS 
COMPANY, Boston,—THOMAS, COWPERTHWAITE & CO., 
PHILADELPHIA,—BELKNAP & HAMERSLEY, Hartrorp,— 
CUSHING & SONS, BALTIMORE,—TRUMAN & SMITH, 
CINCINNATI. 


1838. 


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ATHEMATICS ‘ 
: LIBRARY. 


PREFACE. 


Wiruin the last few years a great change has taken 
place in public sentiment, as to the importance which 
should be given to mathematical studies in comparison 
with other branches of education. Until recently it was 
thought that mere practical rules, unaccompanied by 
demonstration, were abundantly sufficient for all useful 
applications of mathematical science, and that the mind 
of the scholar could find richer nutriment in Virgil and 
Homer, than in the propositions of Euclid, or the sublime 


theories of Newton. 


But it is auspicious to the cause of sound learning, 
that these opinions have given place to more rational 
views of education; that we are at length convinced it 
is better to reason than merely to remember; and that 
the value of an education is to be estimated by the 
ability which it gives to the mind of thinking profoundly 
and reasoning correctly. 


In presenting to the public the following Treatise on 
Shadows and Perspective, the author cannot but flatter 
himself that he shall add something to the common 
stock of useful knowledge. The subjects treated of 


are certainly useful: to the architect and draftsman a 
| A 2 


426240 ‘ 


IV PREFACE. 


knowledge of them is indispensable. To find with ma 
thematical accuracy the lines of shade and shadow on a 
complicated building,—which parts are to be darkened, 
and which parts are to be made light in a drawing of it, 
is certainly a difficult problem unless it be solved on 
scientific principles. 


The art of Perspective teaches us how to represent 
on a surface one or more solid bodies, in such a manner 
that the picture shall exhibit the same appearance as is 
presented by the objects themselves. It is by this art 
that the painter is enabled to present to the eye the 
almost living landscape, with its hills, its valleys, its 
waterfalls, and its rich foliage, varied by the beautiful 
tints of colour, and relieved by alternate light and shade. 
It is this art which has stamped the canvass with the 
intelligence of the human countenance, and caused it to 
be looked upon as the remembrancer of departed worth 
and the record of former times. It is this art which 
presents in a panorama a city in all its proportions, 
and causes the spectator to feel that he almost partici- 
pates in its bustle and business. This art also enlarges 
the pleasures of sight, the sense through which the 
mind receives the most numerous and pleasing impres- 
sions. 


Without perfect accuracy in the perspective, the pro- 
portions of objects cannot be preserved; and the skill 
of the draftsman, or the genius of the painter, is exerted 
in vain, if nature be not correctly copied. 


PREFACE. V 


The manner of finding the shadows of objects, and 
the common methods of perspective, depend on mathe- 
matical principles, and are susceptible of demonstration. 
The want of a work demonstrating those principles 
rigorously, has long been felt, and especially in the 
Military Academy, where the subjects have been taught 
by lecture for several years. If the one now submitted 
be found not to merit the approbation of the public, the 
author hopes it will at least be received with indulgence. 


The author would beg leave to express his acknow- 
ledgments to two or three friends, to whom he is indebted 
for drafts of the diagrams; and also, to repeat his ex- 
pressions of thankfulness to the Cadets for the interest 
they have taken in the work. But for their liberality it 
could not have appeared 


Military Academy, } 
West Point, March, 1832. 


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CONTENTS. 


CHAPTER I. 


DEFINITIONS AND FIRST PRINCIPLES. 


CHAPTER II. 


APPLICATIONS AND CONSTRUCTIONS. 


- To-find the Shadow of a Right Line . . 2... « «1% 
. To find the Shadow of the Abacus and Pillar . . : 
. To find the Shadows of the Chimneys and House . 

- To find the Shade and Shadow of the Cylinder . 

. To find the Ellipse from its Conjugate Diameters . 


To find the Shadow of the Rectangular Abacus ona Golan 


. To find the Shadow of a Cylindrical Abacus on a Column 
- To find the Shadow of a Cylindrical Abacus on a Wall 

. To find the Shadow of the Inverted Frustum ofa Cone . 
. To find the Shade and Shadow of a Sphere . ° 

. Of Brilliant Points. . . boise 

. To find the Shade and Shaner of the Ellipsoid - 

. To find the Shadow on the Interior of the Niche 

. To find the Curve of Shade on the Torus 

. To find the Shadow on a Surface of Revolution 

. To find the Curve of Shade on the Surface of eosin 
. To find the Line separating the dark from the illuminated 


part of a Surface of hey and the Shadow cast by 
the Surface 

To find the Shades atid Shedawe on the ae eo Shaft of 
the Doric Column . 

To find the Shades and ee on ihe Capital of ane Doris 
Column sil Reh an IR pan ngs Sl 

Of the Helicoid . a hes were 

To draw a Tangent Plane to nna Helicoid <e aitany te 

To find the Lines of Shade and Shadow on the SRN 


Page 


61 


Apex 


= OO Ost HD OP 


| oe A ae 


CONTENTS. 


LINEAR PERSPECTIVE 


CHAPTER I. 
Page 
- Definitions and First Principles . : . 95 
- To find the Perspective of a Cube and the Peispactive of its 
Shadow = Ae elie chia Nahe ities Gates 98 
CHAPTER IL. 
. Of the Method of Perspective by Diagonals and Perpendiculars 101 
. To find the Perspective of a System of Cubes and the Per- 
spective of their Shadows ; . 107 
. To find the Perspective of four Peace: - 112 
. The Perspectives of Tangent Lines are tangent to sah aitiek 116 
. To find the Perspective of a Circle - 117 
. To find the Perspective of a Cylinder . : - 119 
. To find the Perspective of the Frustum of an ievortey Gne 123 
. To find the Perspective of a Niche ° - 127 
. To find the Perspective of aSphere . . ... . - 131 
. To find the Perspective of the Groined Arch .. . - 138 
- To find the Perspective of a House ..... . . 147 


* SHADES AND SHADOWS. 


CHAPTER I. 


DEFINITIONS AND FIRST PRINCIPLES. 


1. Licur is that, which proceeding from an object - 
and falling upon the eye, produces the sensation of sight. 

2. Bodies which emit or give out light, such as the 
sun, a candle, &c., are called luminous bodies. 

3. Light is supposed to emanate from every point of 
a luminous body, and to proceed in right limes when not 
deflected or turned from its course. 

4. The right lines, along which the light is supposed 
to move, are called rays of light. 

5. Since light flows from each point of a luminous 
body in every direction, the rays of light flowing from 
a single point form a system of diverging lines. ‘These 
rays are more divergent when the luminous object is 
near; the divergency diminishes when the distance to — 
the luminous object is increased; and becomes nothing, 
or the rays become parallel, when that distance is infinite. 

6. The sun, the chief source of light, is so far dis- 
tant from the earth, that the rays of light coming from 
it may be regarded as parallel. As it is only proposed, 
in this elementary treatise, to find the shades and 
shadows of objects as they appear in nature, the rays of 
light, in the constructions here given, will be considered 
as parallel with each other. 


"ae 


‘ 


10 TREATISE ON 


Having considered the subject under this pomt of 
view, it will be easy to extend the principles developed, 
to embrace the cases in which the rays of light are con- 
vergent or divergent. 

7. In the graphic constructions which follow, it be- 
comes necessary to represent rays of light by right lines. 
When a line is assumed and called a ray of light, it is 
to be understood as merely indicating the direction of 


.the light as it falls upon the body whose shade or shadow 


is considered. 

8. Bodies or objects may be divided into three classes: 

i°. Luminous bodies, or those which give out hight; 
such as the sun, a candle, &c. 

2°, Opaque bodies, or those which intercept the light ; 
such as wood, stone, iron, &c. 

3°. ‘Transparent bodies, or those through which the 
light passes freely; such as water, glass, &c. 

9. Light emanating from a luminous object, and 
fallmg upon an opaque b ><dy, is received by that part of 
the body which is towards the source of light. This 
portion of the surface, on which the light falls, is called 
the wluminated pari of the body. That portion on which 
the light does not fall, is called the shade of the body ; 
and the line which separates the illummated part from 
the shade is called the line of s/.ade. 

Let C (Pl. 1, Fig. 1), a point in the plane of the paper, 
be the centre of a sphere, CB its radius, and AF, also 
in the plane of the paper, a ray of light. The rays of 
light being paralle), and falling upon the sphere in the 
direction AF’, may be regarded as forming a cylinder, 
of which AC 1s the axis. The surface of this cylinder 
is tangent to the sphere in tle great circle of which the 
line BE, drawn at right angles to AC, is the projection. 
—(Davies’ Des. Geom. 108.) 


SHADES AND SHADOWS. 11 


All the rays of light which lie within the surface of this 
cylinder, pierce the sphere in the illuminated part. But 
the sphere being supposed opaque, the light 1s inter- 
cepted, and does not fall on the opposite hemisphere. 
Hence, 
* 1° The surface of the hemisphere BFE, towards the 
source of light, is the illuminated part of the sphere. 

2°. 'The surface of the hemisphere BPE, opposite the 
source of light, 1s the shade of the sphere. 

3°. The great circle of which BE is the projection, is 
the line of shade. 

10. The shadow of an object is that part of space 
from which the object excludes the light of a luminous 
body. Hence, the opaque bodies only cast shadows. 

Referring again to the sphere of which C is the centre, 
and to the cylinder of which AC is the axis, it follows, 
from the above definition, that all the space contained 
within the surface of the cylinder, estimated from BE, 
in the direction CP, 1s but the indefinite shadow cast by 
the sphere. 

11. Let us now suppose a body to be placed at any 
distance from the sphere, and opposite to the source of 
light. It is piain that the sphere will prevent the light 
from falling upon a part of the surface of the body so 
placed.’ The part of the surface from which the light is 
excluded is called the shadow on the body ; and the boun- 
dary of this shadow, is called the kine of shadow. 

T’his shadow on the body is contained within the sur- 
face of the cylinder of rays that is tangent to the sphere, 
and the line of shadow is the line of intersection of the 
surface of this cylinder with the body on which the 
shadow falls. 

12. If a single point, in space, be supposed to inter- 
cept the light, the direction of its shadow will be the 


12 TREATISE ON 


same as that of the ray passing through it; and the 
shadow will lie on the opposite side of the point from 
the source of light. 

13. If the light be intercepted by a right line, the 
shadow will be determined by drawing rays of light 
through all its points. These rays being parallel, and 
intersecting the same right line, are all contained in a 
plane passing through the line so intersected. This 
plane is called a plane of rays. 

14. ‘he shadow of a curved line is determined by 
drawing rays of light through all its points. ‘These rays, 
being a system of parallel lines, may be regarded as 
forming the surface of a cylinder passing through the 
curve. ‘This surface is called the surface of a cylinder 
of rays. 

15. Applying what has been shown of the sphere to 
any opaque body, and recapitulating, in part, what has 
already been said, the following definitions and prin- 
ciples may be laid down. 

1°. The illuminated part of a body is that portion of 
its surface on which the light falls. 

2°. The shade of a body is that part of the surface 
from which the light is excluded by the body itself. 

3°. The line of shade is the line separating the shade 
from the illuminated part of a body, and is the line of 
contact of a cylinder of rays tangent to the body. 

4°. The indefinite shadow of a body is that part of 
space from which the body excludes the light, and is 
limited by the surface of the tangent cylinder of rays. 

5°. The shadow on a body is that portion of its sur- 
face from which the light is excluded by an opaque body, 
between it and the source of light. The boundary of 


this shadow is the line of shadow. 
6°. The line of shadow is the intersection of the sur- 


SHADES AND SHADOWS. 13 


face of a cylinder of rays, tangent to the opaque body, 
with the surface on which the shadow falls. 

7°. The shadow cast by a point upon any surface, ts 
determined by finding where a ray of light drawn 
through the point,pierces the surface. 

8°. The shadow cast by a right line upon any surface 
is the intersection of a plane of rays passing through 
the line with the surface. 

9°. The shadow of a curve upon any surface is the 
intersection of the surface of a cylinder of rays passing 
through the curve, with the surface on which the shadow 
falls. 

10°. The line of shade upon a body is always a 
line of contact, and the line of shadow a line of inter- 
section. 

11°.“Since the line of shadow is determined by the 
surface of the tangent cylinder of rays, it follows, that 
the line of shade is the line which casts the line of 
shadow on the body where the shadow falls. 

12°. When the curve casting the shadow is a plane 
curve, and has such a position that its plane is a plane 
of rays, the rays of light drawn through all its points 
will form a plane and not a cylindrical surface. In such 
case, the shadow of the curve upon a surface is the inter- 
section of its plane with the surface. 

16. If a right line is tangent to a curve in space, the 
shadow of the right line will be tangent to the shadow 
of the curve. For, the plane of rays which determines 
the shadow of the right line will be tangent to the cyl- 
inder of rays which determines the shadow of the curve; 
therefore, their intersections by the surface on which the 
shadows fall are tangent to each other. 

17. If two curves are tangent to each other in space, 
their shadows on any surface will also be tangent. 


ee 
a 


14 TREATISE ON 


for, the cylindrical surfaces which determine the shad- 
ows of the curves are tangent to each other; hence, 
the curves in which they intersect the surface on ‘which 
the shadows fall, are also tangent to each other. 

18. The shade of an object being considerably 
darker than the illumimated part, is easily distinguished 
from it; and the line of shade is, in general, distinctly 
marked. Inthe drawings, the shade will be distinguished 
by small parallel lines, as in the hemisphere EPB (PI. 
1, Fig. 1). 

Shadows also, appearing darker than that part of the 
same surface on which the shadow does not fall, will, in 
the drawings, be darkened by small parallel lines, in the 
same manner as the shade. 

19. Inthe drawings to be made, two planes of projec- 
tion will be used, as in Descriptive Geometry. 

20. Since the projections of an object should rep- 
resent the appearance which the object itself presents to 
the eye, situated at an infinite distance from, and ina 
perpendicular to, the plane on which the projection is 
made, only that part of its surface must be shaded in 
the drawing which is in the shade, and seen by the eye. 
The portion of the surface which is in the shade and not 
seen by the eye, must not be darkened in the drawing; 
if it were, the drawing would not be a true representa- 
tion of the object. 

In hike manner, a shadow upon one of the planes of 
projection, or upon any other surface, may be concealed 
from the eye, in which case it 1s not to be represented in 
the drawing. 

21. ‘The same general rules are observed in making 
the projections of bodies as in Descriptive Geometry. 
All the bounding lines which are seen are made full: » 
and all auxiliary lines either dotted or broken. 


SHADES AND SHADOWS. 15 


‘The rays of light will be represented by small broken 
lines. | 

22. To enable us to determine the shade and shadow 
of an object, there must be given, 

1% The position of the opaque object casting the 
shadow, which is determined when its projections are 
given. 

2°. The surface on which the shadow falls. 

3°. The direction of the light. 

From these data, the shade and shadow of any object 
can always be found. 

The terms plan and elevation, are generally used by 
Architects instead of the terms, hortzontal-projection and 
vertical projection. As the terms are synonymous, either 
may be used. 

In architectural drawings it is customary to suppose 
the source of light on the left of the object, and the 
rays to have such a direction that their horizontal and 
vertical projections shall make angles of 45° with the 
ground line. When this is the case, it is plain that the 
rays will be parallel to the diagonal of a cube whose 
faces are parallel and perpendicular to the planes of 
projection; and since the diagonal of a cube makes an 
angle of 35° 16’ with the plane of cither of its faces, 
it follows that any ray of light must make an angle of 
35° 16 with the planes of projection, when both its pro- 
jections make angles of 45° with the ground line. 


16 TREATISE ON 


CHAPTER II. 


APPLICATIONS AND CONSTRUCTIONS. 


PROBLEM I. 


To find the shadow cast by a right line upon the horizontal 
plane of projection. 


23. Let IL (PI. 1, Fig. 2) be the ground line, (BD, 
B'D’) be the given line; (A,A’) a ray of light, A being 
its horizontal, and A’ its vertical projection. 

Through any point of the line,as (B, B’), conceive a 
ray of light to be drawn. _ Its projections will be respec- 
tively parallel to A and A’, and the ray will pierce the 
horizontal plane at E, which will be one point in the 
shadow of the right line. 

Through any other point of the nght line,as (D, D’) 
conceive a ray of light to be drawn. Its projections will 
also be parallel, respectively, to A and A’, and the ray 
will pierce the horizontal plane at F, which is a second 
point in the shadow of the given line. 

Since the shadow of a right line upon a plane is a 
right line, the line EF is the shadow cast upon the hori- 
zontal plane by the Ime (BD, b’D’). The indefinite 
right line HG is the indefinite shadow on the horizontal 
plane, cast by an indefinite right line passing through the 
two points (B,B’) and (D,D"’). It is plain that this shadow 
HG, is the horizontal trace of a plane of sisi passing 
through the line (BD, B’D’). 

The right line (BD, B’D’) may be so situated that the 


SHADES AND SHADOWS. 17 


whole, or a part of its shadow, will fall on the vertical 
plane. If it be required to find that shadow, we have 
only to construct the vertical trace of the plane of rays 
passing through the given line. 

It is also apparent, that the pomt H, in whieh the line 
(BD, B'D’) produced, pierces the horizontal plane, is in 
the right line joming the points E. and F’, since the three 
points are all in the trace of the same plane. 

Hence we conclude, that the point in which a right line 
pierces a plane ts one point_of the indefinite shadow of the line 
upon the plane. And, extending the principle, the indefinite 
shadow cast by aright lne upon any surface, passes through 
the point inwhich the right line, produced if necessary, prerces 
the surface. 

24. When a right line is parallel to the plane on which 
the shadow falls, the shadow will be parallel to the line 
itself. For if the line and its shadow be not parallel, 
they would, if produced, intersect, which they cannot do, 
since a line cannot intersect a plane to which it is parallel. 

Construction of the figure. Draw on the paper the inde- 
finite line [L, to represent the intersection of the planes 
of projection, or the ground line. Then draw the lines 
BD, B'D’ to represent the projections of the given line; 
also, A and A’ to represent the projections of the ray of 
light. This being done, through B and D, draw lines 
parallel to A. ‘Through B’ and D’ draw lines parallel to 
A’. At the pomts E’ and EF’, erect in the horizontal 
plane perpendiculars to the ground line, and mark the 
points E and F' in which they meet the parallels first 
drawn. Draw the line EF’, which is the shadow required 


18 ; TREATISE ON 


PROBLEM II. 


Having a rectangular pillar with a rectangular abacus 

placed upon tts upper face ; wt ts required to find the shadow 

“cast by the abacus on the faces of the pillar, and the shadow 
cast by the abacus and pillar on the horizontal plane. 


z 

25. Let the square ab (PILI. Fig. 3) be the horizontal 
projection of the pillar, cd its vertical projection, EC the 
horizontal projection of the abacus, and EG its vertical 
projection. The direction of the light is shown by the 
ray (A, A’). 

Let us first find the shadow of the line (BC, FG), on 
the front face of the pillar. Through (B, F) a pomt 
of the line, conceive a ray of light to be drawn; the 
two projections are respectively parallel to A and A’. 
The point (h,h’), in which this ray pierces the front face 
of the pillar, is one point in the shadow of the line (BC, 
FG) on that face. But the line (BC, FG) being par- 
allel to the front face, the shadow is parallel to the line 
itself; hence hm, drawn parallel to the ground line, is 
thevertical projection of the required shadow. ‘The 
horizontal projection is in the line Ad. 

The line of which BE is the horizontal projection, 
and which is vertically projected at F, casts a shadow 
upon the front face of the pillar, on the face na, and on 
the horizontal plane. 

Conceive a plane of rays to be passed through this 
line—this plane is perpendicular to the vertical plane 
of projection, and its vertical trace Fuh, is parallel to 
the vertical projection of the light; and n’h’ is also the 
projection of its intersection with the front face of the 
pillar: therefore nh’ is the vertical projection of the 


he 


SHADES AND SHADOWS. . 19 


shadow cast on the front face of the pillar by a part of 
the line (BE, F)\—the horizontal projection of the same’ 
shadow is xh. Drawing through n,the line pn parallel 
to the horizontal projection of the light, gives Bp the 
horizontal projection of that part se thes line which 
casts a shadow on the front face. ; 

The part of the line from p to q,casts a right line of 
shadow on the face na, which shadow is vertically pro- 
jected atn’. The part gE casts a shadow on the horizon- 
tal plane. In finding the shadow of the abacus on the 
horizontal plane, we will begin with the point (C, D’). 

Through this point suppose a ray of light to be drawn. 
* Such ray will pierce the horizontal plane at H. Through 
(D, D’) conceive a ray also to be drawn. Such ray will 
pierce the horizontal plane at I. Hence HI is the 
shadow cast upon the horizontal plane by the line 
(CD, D’). 

But Lis also the shadow on the -horizontal plane of 
one point of the line (ED, E’D’); and since this line is 
parallel to the plane, IL drawn parallel to ED, is its 
indefinite shadow. ‘The shadow is limited by the line 
EL, drawn through E, parallel to the horizontal pro- 
jection of the hght. 

Through the point (C, G) conceive a ray of light to 
be drawn. It pierces the horizontal plane at 7. The 
line Hz is the shadow cast by the vertical line (C,GD’) 
of the abacus. 

Through the point z,draw 7K parallel to BC, and pro- 
duce it till it meets £K, drawn through 4, parallel to the 
horizontal projection of the light. Then K7is the shadow 
cast by (kC, &’G),a part of the front and lower line of 
the abacus. The ray of light through the point (4, #) 
touches the edge of the piles at m, arf pierces the hori- 
zontal plane at K. 


20 TREATISE ON 


From K to & the shadow is cast by the part of the 
edge (6, md). 

Returning to the point L, Lf is the shadow cast by the 
line (E, E’F), fy the shadow cast by the line (Eg, F’), and 
ga the shadow cast by that part of the edge between n’ 
and the horizontal plane. 

The light falls on three faces of the abacus, viz: the 
upper face, whose vertical projection is the hne E’D’; 
the side face whose vertical projection is the line ET; 
and the front face whose vertical projection is the rec- 
tangle FD’. The three remaining faces are in the shade, 
but neither of them is seen in either projection. 

The part of the front face of the pillar on which the 
shadow of the abacus falls, and the shadow of the pil- 
lar and abacus on the horizontal plane, are shaded. 
The part of the shadow on the horizontal plane which 
falls under the abacus cannot be seen in the horizontal 
projection, and is therefore not shaded. 

It may not be out of place to remark, that the lines 
of shadow which have been determined, are but the in- 
tersections of planes of rays passing through the lines 
casting the shadows, with those planes on which the 
shadows fall. 

Thus h'm is the trace, on the front face of the pillar, 
of a plane of rays passing through (BC, FG). The 
shadow wh’ is the trace, on the front face of the pillar, 
of a plane of rays passing through the line (EB, F); HI 
is the trace, on the horizontal plane, of a plane of rays 
passing through (CD, D’); Hz is in the horizontal trace of 
a vertical plane of rays passing through the vertical line 
(C,D’G). The shadow 7K is in the horizontal trace of 
a plane of rays passing through (BC, FG). The shadow 
Ké is in the horizontal trace of the vertical plane of 
rays passing through the edge (6,dm); and this plane 


SHADES AND SHADOWS. 2] 


cuts off the part (£C, /’G) of the line (BC, FG), which 
casts the shadow Kz on the horizontal plane. Similar 
explanations may be given of the other lines of shadow 
on the horizontal plane. 


PROBLEM IIL. 


It ts required to find the shadows cast by the cornices of 
the chimneys of a house, on the faces of the chimneys ; the 
shadows cast by the cornices and chimneys on the roof of the 
house, and the shadows cast by the roof on the walls, and on 
a horizontal plane at a given distance below the eaves. 


26. Let the rectangle BCEA (PI. 2) be the horizontal 
projection of the outer lines of the eave-trough. These 
lines are contained in a horizontal plane, and are ver- 
tically projected in the line A’C’.. The lines of the inner 
rectangle are the intersections of the outer faces of 
the walls with the horizontal plane; they are made 
broken in projection, being concealed by the roof. The 
lines of the middle rectangle are the horizontal pro- 
jections of the eaves of the roof. ‘The eaves are 
supposed to be in the same horizontal plane with the 
upper line of the eave-trough, and are therefore verti- 
cally projected in the line AC. 

The line in which the side roofs intersect each other, 
and the lines in which the side roofs intersect the end 
roofs, are represented by full lines in horizontal pro- 
jection. 

With regard to the chimneys, the lines of the outer 
rectangle, as 7/hg, are the horizontal projections of the 
outer lines of the cornice; the lines of the middle 
rectangle are the intersections of the outer faces of the 


22 ; " TREATISE ON 


, 


chimneys with the roofs; and the inner rectangle is the 


' projection of the flue or hollow part of the chimney. 


K. 


The vertical projections of the roofs ‘and chimneys can 
be understood from the figure, excepting, perhaps, the 


manner of determining the lines in which the front and .- 


back faces of the chimney in the front roof, intersect the 
roof. After the horizontal projection of the chimney is 
made, these lines are thus found: 

The faces of the chimney intersect the front roof in 


Imes parallel to the ground line. * Producing the hon- 
zontal projection of the front face till it meets mn G, the 


‘line of the.end roof, we obtain IG, the horizontal pro- 


_ jection,of the line of intersection. But G is vertically 


* projected in G’, therefore GT drawn through G’ parallel 


to the ground line, is the vertical projection of the hlne 


“in which the front face of the chimney intersects the 


roof. For similar reasons, the line drawn through H’ 
parallel to the ground line, is the vertical projection of 
the line in which the back lace of the chimney intersects 
the roof. 


~ In finding the shadows, let us begin with the chimney 


of the end roof. 


~ Through (/,/”"), a point of the lower line of the cor- 


nice, conceivea ray of light to be drawn. It pierces the 
front face of the atone in the point (g, 9’); and the line 
g'p is the vertical projection of the shadow cast by (fh, 
f‘h’) on the front face ofthe chimney. The line og’ is the 
vertical projection of the shadow cast on the front face 
by a part of the line (fc. f”). A similar construction gives 


the shadow cast on the front face of the other chimney. — 


To upd the sna which the chimneys cast on the 
roofs. , 

Through the line (/q, i) conceive a plane of rays to 
be passed. Smee the line (hg, h’) is perpendicular to 


"> 


“ Fea 


BRA 


ae 


Pgs: 


D 


GW ce ert 


K’ 


SHADES AND SHADOWS. oF 


the vertical plane of projection, the plane of rays through 
it will also be perpendicular to the vertical plane; and 
hence its vertical trace will be the line h’'Z'F", parallel to 
the vertical projection ofthe light. This plane ofrays in- 
tersects the horizontal plane of the eaves in a line which 
is perpendicular to the vertical plane at F”, and of which 

I’, perpendicular to the ground line, is the horizontal 
projection. It also cuts the line of intersection of the 
side roofs in a point whose vertical projection is /’ and. 
whose horizontal projection is £; and hence LF’, kf’ are 
the horizontal] projections of the intersections of the 
plane of rays through (hq, h’) with the side roofs; which 
intersections are the indefinite shadows cast by (hq, h’) 
on these roofs. ‘The shadows are limited in horizontal 
projection by the lines hn, and gm,drawn parallel to the 
horizontal projection of the light. Therefore kn and 
km are the horizontal projections of the shadows cast 
on the side roofs by the line (hq, h’). 

It may be here remarked, that all planes of rays which are 
perpendicular to the vertical plane of projection are parallel 
to each other, and will consequently intersect the side roofs in 
lines parallel to k¥" and kF. 

Drawing through m the line ms, parallel to the ground 
line, and through 2, 7s parallel to the horizontal projec- 


tion of the light, gives ms for the shadow of the lne | 


(iq, fh’) on the side roof. ‘The line us,is the horizontal 
projection*of the shadow cast by the perpendicular (2, 
ff’). This shadow is in the trace of a vertical plane 
of rays through (2, ff"), and is limited at u by the intersec- 
tion of a plane of rays through (/7, /”) with the side roof. 

The line wv, lying in this intersection, is the shadow 
cast bya part of the line (f7,f") on the side roof. The 
shadow at v is Cast by that point of the back edge of the 
chimney which is vertically projected at o. 


JA TREATISE. ON 


From v the shadow is cast by the back edge of the 
chimney. 

Returning to the shadow cast by the point (A, /), we 
find, first, the shadow ng,-which is-cast by the perpen- 
dicular (h, Wh"), and then draw through « a parallel to 
the ground line; this parallel i is the indefinite shadow 
cast by the line (fh, /’h’) on the side roof. We next 
find the point in which a ray of light through the point 
whose vertical projection is p, pierces the end roof, 
which is at (¢,/). Joining the point ¢ with the point in 
which the parallel through x meets the intersection of 
the side and end roofs, gives the shadow cast by the line 
(fh, fh’) on the end roof. From ¢ the shadow is cast 
by the edge of the chimney. 

We find the vertical projection of this shadow by pro- 
jecting its points into the vertical traces of the planes 
to which they respectively belong. ‘Thus m is verti- 
cally projected at ms ats,uatu,vatv,y at 7, k at 
kijnatn,and «at 2’. That part of the shadow which 
is on the front roof is all that can be seen in vertical 
projection. The boundary of that on the back roof 
is therefore dotted. 

The shadows of the other chimney are found ina 
manner so entirely similar as hot to require a particular 
explanation. 

Let us now find the shadows on the walls of the house. 

Through (BC, ac), the upper line of the convex part 
of the eave-trough, conceive a plane of rays to be passed. 
Its trace on the front wall limits the shadow cast by the 
eave-trough. The point (6, 6’) is in the trace of this 
plane, and 6’e isthe required line of shadow. The shadow 
b'z is cast by the corresponding line of the end eave- 
trough, which also casts a. shadow on the end wall that is 
vertically projected at z. 


«& 


SHADES“ AND SHADOWS. 25 


- To find the shadow cast brine house on the horizontal 
Alar of projection. 

Through (CE, C’), the upper ‘line of the end eave- 
trough, conceive a plane of rays. to be passed. ‘The line 
LE is its trace on the horizontal plane, and consequently 
the shadow of the line ( 

The small vertical Tine (GC) casts its shadow Le’ 
in the line CL. » Through ¢ draw cd parallel to the 
“ ground line, and ities it till H meets Dd drawn par- 
allel to the horizontal projection of the light.” Then 
dc’ is the shadow cast by the le (DC,D’c), and the ray 
through (D, D’) touches the corner of the house at e, 
and pierces the horizontal plane at d. 

The line aE’, drawn through FE’ parallel to the ground 
line, and limited by Aa’ drawn parallel to the horizontal 
projection of the light, is the shadow cast on the hon- 


« zontal plane by the line (AE, A’C’). The line (A, aA’) 


casts the shadow a'r; a part of the line (BA, a) casts the 
shadow rr: from 7’ the line of shadow is cast by the 
corner of the house below z. 

The shadow on the horizontal plane which lies under 
the eave-trough and cornice, is not seen in Bs os 
projection. 


* 


PROBLEM IV. 


Having ‘given an oblique cylinder, and the direction of the 
hight, it 1s required to find the shade on the exterior surface, 
the shadow of the upper circle on the interior surface, and. the 
shadow of the cylinder on the horizontal plane. 


27. Let (AB, A’B’), (PI. 3. Fig. 1), be the axis of the 
cylinder, and suppose the projections to be made as 
seen in the figure. 


af AM tiles 2 Qe OR ee 
= ‘ the : ae a "% me i nih 
Mp * ®. > . at 

bg o is * . ot 
og ‘ t 7 «’ * ¥s ; 

f ae : Mane aa 

= ths - 28 OY ans raga TSH ON * 

- ‘. If we suppose two abe planes of rays to be drawn: — 

pe to the surface of the eylifider, the elements of contact 

. . will separate the dark from the illuminated part of the. 

me -.» surface, and will consequently be lines of shade. 

yf In order to pass these planes, we first-pass aplane of. - 

*. rays through the axis of the cylinder, to which the 

ha ~- tangent planes will both be parallel. | 

“* Through (B,B’), a point of the axis, conceive arayof _ . 


light to be drawn—it pierces the horizontal plane at C. 
The avis of the cylinder pierces the horizontal plane at — 
A—therefore AC is the horizontal trace of a plane of - 
rays passing through the axis of the cylinder. 

‘Let two'lines be now drawn tangent to the base of the 
cylinder, and parailel to AC. ‘These lines are De and 
Gf", and aré also the horizontal traces of the planes 
of rays tangent to, the surface of the cylinder. The 
- elements of contact pierce the horizontal plane at D and 
~G. Hence, DE and GF, drawn parallel to AB, are the 
horizontal projections of the elements of contact: and 
by projecting D and G into the ground line at D’ andG’, 
and E and F into the vertical projection of the upper 
circle at E’ and F’, we determine D’E’ and G’F’, the 
vertical projections of the elements*of contact. The ~* 
points D, A and G,are in the same straight line perpen- 
dicular to AC; and DHG is a semicircle. 
| The light falls'on the half of the exterior surface cor-, 
vs responding to the semicircle DHG—the remaining half | 
- is inthe shade. Thelines (DE, D'E’,)\(GF, GF’) are. 
lines of shade, and being also elements of the surface, 
are generally called elements of shade. 

In the horizontal projection, we do not see the i & 

» which is on the under surface of the cylinder—therefore, 
we darken only that part which is included between the” 
elements DE and LN. In the vertical projection, we 


Legs 
oe 


wre 


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| 
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. 


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et BOGE pie 
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<_< = 4 
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L? ITC Aal 
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——* es oe = - a ete a GIRL NOU ite 9 Senor Ie hE Bi = aneteded ET: aay . —_ en. — 


os 


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“wes, 


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Pi 


+ 


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a 8 & ™ Cet f Ae ge ” 
- Ps + ¢ ae * + 
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di a + ~~, + i 
7 4 » , 7, ¥ 4 *» 
, a’ + ¢ 
$s. ; . ™~. 
© eB - « ! a a 
* a 
a . 
SHADES AND SHADOWS. : %. 27 


see that part of the shade included between the element 
G’F’, and the element P’K’, only that portion of the 
surface, therefore, is shaded. ‘2 
To find the shadow cast by the upper circle on the 
interior surface. . 
This shadow begins at the points (4, E’), EE) in 
which the elements of shade meet the upper circle, and 


is cast by the semicircle whose horizontal projection is. 


dak. 

If the surface be intersected by planes of rays parallel 
to the axis of the cylinder, each plane so drawn will in- 
- tersect the surface in two rectilinear elements. If then, 
through the upper extremity of the element towards the 
source of light, a ray be drawn, it will be contained in 


the plane of rays, and will, consequently, mtersect; the © 


other element lying in that plane—the point of inter- 
section is a point of shadow on the: interior surface. 
But the horizontal traces of all planes of rays parallel 
to the axis of the cylinder are parallel to AC, the 
trace of the’plane of rays through the axis. Therefore, 
* draw any line, as IP, parallel to AC, to represent the 


». trace of a plane of rays parallel to the axis. ‘This plane, 


intersects the surface in two elements, whose horizontal 
projections are Id and PK. ‘Through d, the upper ex- 
tremity of the element towards the source of light, 
draw df parallel to the horizontal projection of the light 
—the point f, in which it meets.PK, is the horizontal pro- 
jection of a point. of shadow on the mner surface. 
The vert.cal projection of this point is found by pro- 
jecting P into the ground line, drawing the vertical, pro- 
jection of the element on which the shadow falls, and 
noting its intersection with the ‘perpendicular to the 
ground line, drawn through the point f; or,by pro- 
jecting the point dinto the upper base, drawing through 

ae « ; a 

. - + : a =e ” 


23 ; TREATISE ON @ 


d' the vertical anaes of a ray of hght, and noting 
its intersection f’ with the vertical projection of the 
element on which the shadow falls. 

If it be required to find the shadow on any particular 
element, as the one, for example, whose horizontal pro- 
jection is LN, we have merely to pass a plane of rays 
through the clement, and determine the other element 
towards the source of light, in which it intersects the 
surface. ‘Then, through the upper extremity of the 
element casting the shadow, draw a ray of light, and 
where it meets the given element, is the penned point 
of the shadow. If we take the element whose “hori- 
zontal projection, is LN, Riis the horizontal trace of 
the plane of rays passing through it—the element 
towards the source of light, in which this plane inter- 
sects the surface, is horizontally projected*in Ré, and g 
is the horizontal projection of the pointof shadow cast 
by its upper extremity. | 
| At g, the horizontal projection of the curve of 

shadow is tangent to the line LN: for the vertical plane . 
of which LN is the horizontal trace, is tangent to the ‘ 
vertical cylinder, which projects the curve ef shadow, * 
along the element that pierces the horizontal plane at g. 

The lowest point of the curve of shadow, or that 
point which is farthest from the upper circle, is found by 
passing a plane of rays through the axis. For, since all 
the rays of hight make equal angles with the horizontal] 
plane, or with the plane of the upper base, the distance 
of the points of shadow below the upper base will 
increase, as the distance between the element casting 
and the element receiving the shadow, is increased. But, 
the plane of rays through the axis intersects the surface 
in elements farther from each other than the elements cut 
out by any other of the parallel planes of rays. Hence, 


? 
* ° ”* 4 i ; f 
» &? - 
bral 
Bis 
SHADES AND SHADOWS. 29 


it determines the lowest point’ of shadow, which is 
(A, h’). ig 

Having found as many points of the shadow as may 
be necessary, let the projections of the curve be accu- 
rately described. . 

In horizontal projection, the curve of shadow is seen 
from fF until it crosses the horizontal projection of the 
upper circle, thence to the point Eit is concealed by 
the surface of the cylinder. The part of the inner sur- 
face lying above the curve of shadow, receives the light ; 
the part lying below it is in the shadow. That part of 
the surface which is in the shadow, and seen, is shaded 
inthe drawing. With respect to the vertical projection, 
no part of the curve of shadow is seen, since it lies en- 
tirely on the inner surface of the cylinder. . 

To find the shadow of the cylinder on the horizontal 
plane. — 7 

“The lines 6f the cylinder which cast shadows on the 
horizontal plane are, the two elements of shade, and , 
the half of the upper circle which is opposite the source | 


» of light, and of which ENF is the horizontal pro- 
jection. | 


The shadows cast by the elements of shade fall in the 
horizontal traces of the tangent planes of rays—hence, 
we may consider De and Gf”, as the indefinite shadows 
of the elements of shade. 

If through the upper eircle of the cylinder we con- 
ceive a cylinder of rays to be passed, its axis will be 
a ray of light passing through the centre’(B, B’), and 
will pierce the horizontal plane at C. But, since the 
plane of the upper circle and the horizontal plane are 
parallel, they will intersect the surface of the cylinder 
of rays in equal circles. Therefore, a circle described 
with C as a centre, and a radius equal to BK’, will be the 


® w 


* 


zc”)!lU OS ee iy +" ai a + ™ a 
+ + ri ei 
4 =. ’ 
- ¥ é ¥ % 
a" .” re 
. agi . Bee, * 
oe ‘ a ? 
A é 
il %.. * * 
‘ . 7 a b 
30 TREATISE ON 


shadow cast by the upper circle on the horizontal plane. 
The shadows cast by the elements of shade will be 
tangent to this circle of shadow at the points f” and e; 
‘which points are found by drawing a line through C per- 
pendicular to AC, or by drawing the horizontal projec- 
_tions of rays through the points F and Ey 
’ ” Although we have spoken of the shadow-cast on the 
horizontal plane bythe whole of the upper circle, yet it 
is obvious that the semicircle which is towards the source 
of light, and which casts a shadow on the inner surface 
of the cylinder, cannot cast a shadow on the hori- 
zontal plane, unless we suppose the-surface to be trans- 
parent, so'that the light may not be intercepted by it. 
| The part of the shadow on the horizontal plane 
‘which is concealed by the horizontal projection of the 
rface, is not shaded in the drawing. 5 
28. The shadow of a circleyor iddeadl of any curve, 
* ona plane to which it is parallel, is an equal circle Oe 
curve. But*the shadow of a cirele on a piane to which 
it 1s it parallel, is, in general, an ellipse. The shadow ~ «« : 
cast by the centre of the circle isthe centre of the ellipse, »’ 
and the shadow cast by any diameter of thé circle is a 7 
diameter of the* ellip e.» Two diameters of an ellipse ™ 
are said to be conjugate when either of them is parallel» 
to tangent lines drawn through the vertices of the other. ae 
If any two diameters be taken tn a circle at right angles to rat: 
each other, their shadows will be conjugate diameters of the € 
ellipse of shadob. bie 3 
For, if through the vertices of either diameter, tangent 
_lines be drawn to the circle, they will be parallel to the 
other diameter; hence, their shadows on any plane will 
“be parallel to the shadow, of this latter diameter. But 
the shadows of the tangents will be tangent to the | 
shadow of the circle (16), that is, to the ellipse of . oe 
. » 
es. 


in, me *. 


=| 


etme .¥ 


| Fefough 0’. - ey 


SHADES AND» SHADOWS. 31 


shadow. Therefore, the shadow of ‘one diameter is 
parallel to the tangents drawn through thé vertices of 
the shadow of dhe other. Hence the shadows cast by 
two diameters of a circle at right angles to each other, 
are conjugate diameters of the ellipse'o 8 hadow.” - 

29. Itis,oftemrequired to construct ellipses when their « 
conjugate diameters only are known. The construction 
is easiest made by finding the axes Fi the curve. * 

We shall, therefore, give a problem, the principles of 
which are found inCr ouets Conic Sections, for finding the 
axes of an ellipse when two of its conjugate diameters are given. 

Let AB and DS (P1.3, Fig.2), be two conjugate diame- 
ters of an ellipse. Ehren either vertex, as D, of either 
diameter, draw a parallel EF to the other diameter. At 


D erect in a plane perpendicular to AOD, DO’ perpen- . 


dicular to EF, and make it equal to half the parallel 
diameter AB. With O' as a centre, and radius O'D, 
describe the circle CaDb. 


Now we may regard the ellipse whose conjugate di- _ 


ameters are AB and DS, as the shadow of the Byele 
‘Cabo. 

~ The ray of ight through the centre O’ pierces the 
-horizontal plane at O; the diameter ad casts the shadow 


» AB, and the diameter CD the shadow DS. 


It is now required to find two diameters of the circle 


CaDb at right angles to each other, whose shadows _ 


shall also be at. right angles ; ‘ for) 'the conjugate 
diameters of an ellipse, which are perpendicular to each 
other, are the axes. Seg 

To find these diameters it is necessary to construct 


two semicircles having a common diameter in the line | « 


F, one lying in the horizontal plane and passing 
ugh O, the other in the vertical plane and passing 


: ad ; 
* = 


se 


~s * 
*, 


t 


32 TREATISE ON 


Produce O'Dand make DP equal to it. Jom OP and 
bisect it by a perpendicular line—the point N, where the 
perpendicular meets EF’, 1s the common centre of the 
two semicircles. Let them be described with the radius 
NO, or NO”. Then draw the radii O’gE and OUF. 
These radii are at right angles to each other; and so 
are their shadows EO and FO. Hence,GO and LO are 
the semi-axes of the ellipse. ‘The extremities G and L 
are determined by drawing rays through g and s the ex- 
tremities of the perpendicular radii. ‘. 


PROBLEM V. 


Lo find the shadow of a rectangular abacus on a a cyline 
drical column; and also, the shade os the column. 


30. Let the semicircle z SP (Pl. 3, Fig. 3) be the plan 
of a semi-column, and op'd¢' its elevation ; the rectangle 
ep the plan of the abacus, and ¢’d’ its elevation. 

Through the lower line (cz,c’) of the abacus, conceive 
a plane of rays to be passed. This plane will be per- 
pendicular to the vertical plane of projection, and its 
intersection with the surface of the column will be the. 

* shadow cast by the line (cz, ¢’), (15). ‘ 
‘Through ¢ draw c? parallel to the vertical pro- - 
. jection of the light. This line is the vertical trace of 
the plane of rays passed through the line (cz, ¢), and it 
is also the vertical projection of the indefinite shadow 
“east on the column. (Des. Geom. 82.) 

But the ray of light through the point (¢, ¢’) pierces 
the surface of the column at ‘Cre therefore ¢2’ is the 
vertical projection of the shadow cast on the column by 
_ the line (cz, ¢’). The shadow itself 1s an ellipse, and is. 
PA projected in the are zz. aa 


SHADES AND SHADOWS. 33 


To find the shadow cast by the line (cd, cd’): 

The shadow cast by the point (¢,c’) is already found 
at (2,0’). 

Let the surface of the column be intersected by 
planes of rays perpendicular to the horizontal plane. 
These planes will cut the line (ed, cd’) of the abacus in 
points, and the surface of the column in right lines— 
then, drawing through the points of the abacus,rays of 
light, where they meet the elements of the column are 
points of the shadow. 

The line 64, drawn parallel to the horizontal projection 
of the light, is the horizontal trace of a vertical plane 
of rays, and this plane determines (f, k’) a point of the 
shadow. The points (4,7) (g, 2") (g, q) and (n, x’) are 
found in a similar manner. 

The vertical plane of rays, whose horizontal trace 1s 
EF, is tangent to the column. The ray, therefore, 
through the point (14, E’,) touches the column at (a, 7’). 
At this point the line of shadow terminates, and the line 
of shade begins and passes down the column vertically, 
being the element in which the plane EF is tangent to 
the column. 

The part of the surface of the column which is above 
the line of shadow, and the portion which is in the 
shade, are shade in the drawing. 


PROBLEM VI. 


To find the shadow cast by a cylindrical abacus on a eylin- 


drical column ; and the shade on the abacus and column. 


31. Let nsr (Pl. 6, Fig. 2) be the plan of a semi- 
column, fr’ its elevation—cemd the plan of the abacus, 
and c’d its elevation. 


Let the surface of the column be intersected by ver- 
C 


34 TREATISE ON 


tical planes ‘of rays. These planes will intersect the 
lower circle of the abacus, which casts the-lime of 
shadow on the column, in points, and the surface of the 


‘column in right lines. Through the points of the 
abacus let rays of light be drawn—the points in 


which they intersect the elements of the column are 
points of the line of shadow. The plane of. rays, 


whose horizontal trace is hn, parallel to the horizontal 


projection of the light, determines the point of shadow 
(n, n); and the points of shadow (0, 0), (9,q) (8, s) 


(t, ) (u, w) and (¥, v’) are found in a similar manner. 


The vertical plane of rays, whose horizontal trace is 
mv, is tangent to the column along the element of shade. 
The shadow ends and the shade begins at the point (v,v’). 

The shade on the abacus is found by drawing a plane 


of rays which shall be tangent to it—the line p drawn 


tangent to cmd and parallel to the horizontal projection | 
of the rays of light, is the horizontal trace of the — 
tangent plane—the element of contact is the element 


of shade. 


The part of the surface above the line of shadow, as 
well as that portion of it which is in the shade, is shaded 
in the drawing. 


PROBLEM VII. 


Having given a cylindrical abacus and the direction of the — 
light, wt ws required to find the shadow of the abacus on a 
vertical plane, or wall. 


€ 


32. Let anbp (Pl. 6 Fig. 3) be the. plan, and a’cd'b 


the elevation of the abacus, the abacus touching the 


vertical plane on which the shadow falls,in the element 
(p, 70). 


< 


‘3S . get = Plate o. 

‘es, "4 
| 
| 


in 


Hl 
ii 


* 
2 : ‘ 
; + + 4 
: . é # 
. = 2 = 
= | 
: ae = ae ee eee, a. = ny © EB Prudhomme fc | 
OE LL : 2 - = 2 —— —_ - = Se eee, 2 me -- : 


t, . * = —* 


SHADES AND SHADOWS. 30. 


Suppose two squares to be constructed, the one cir- 
cumscribing the upper circle of the abacus, the other 
the lower. The square cdgl is the projection of both 
the squares on the horizontal plane. 

Let two planes of rays be drawn tangent to the abacus. 
The lines ¢¢” and «”, drawn parallel to the horizontal 
projection of the ray of light and tangent to the circle 
nbpa, are their traces; and (t, ¢s), (¢, yx) are the 
elements of contact, and consequently the elements of 
shade (27). 

Now the lines of the abacus which cast lines of 
shadow on the vertical plane, are, Ist, the upper sem1- 
circle (tb, bt'y); 2dly, the two elements of shade (/, ¢'s) 
(7, yx); and 3dly, the lower semicircle (cant, c'xs). 

Through (z,7’), the centre of the upper circle of the 
abacus, let a ray of light be drawn,—the point 2’, in 
which it pierces the vertical plane, is the centre of the 
ellipse of shadow cast by the upper circle. The ray 
through (n, 7’) pierces the vertical at p, hence n'z'p is 
the Neonat cast by the diameter (np, n’). Drawing 
rays through (6, 6’) and (a, a’) determines w and v, the 
extremities of the shadow cast by the diameter (ad, 
ab’). Hence n'p and vw are conjugate diameters of 
the ellipse of shadow (28). 

Drawing rays of hght through the points (d, 6’) and 


(c, a ),determines Of the shadow cast by the tangent ; 


(dg, 6’); also ef the shadow cast by the tangent (cd, 
ab’), and ae the shadow cast by the tangent (cl, a’). 
But since these lines are all tangent to the upper circle ° 
of the abacus, their shadows will be tangent to the 
ellipse of shadow (16); and the same may be shown of 
the lower circle of the abacus and its tangents. | 

The ellipse of shadow cast by the lower circle of the 
abacus is easily found. The point 2” is its centre, og, 


“ 


36 TREATISE ON 


z'y are its conjugate diameters, and the lines dh, dc, 
cg, and gh are tangent to it. The elements of shade 
cast the shadows ¢’u and 7/. Hence, the lines of shadow 
on the vertical plane, are—the semi-ellipse dz qu, cast 
by the lower semicircle (cant, c'xs); the right line wt’, 
cast by the element (¢, ¢s); the semi-ellipse fwn'7’, 
cast by the upper seimicircle (2, ty), and the right line 
#1 cast by the element of shade (2, yz). 


PROBLEM VIII. 


Having given the frustum of an inverted cone, tt 1s re~ 
quired to find the shadow cast by the upper circle on the tnner 
surface, and the shadow cast by the frustum on the plane of the 
lower base. 


33. Let the circle CLHD (P1.4) be the horizontal pro- 
jection of the upper circle of the frustum, and G’H’ its 
vertical projection. Let the circle described with the 
centre A and radius AK be the horizontal, and I’K’ the 
vertical projection of the lower base. The axis of the 
cone being supposed perpendicular to the horizontal 
plane, is projected on the horizontal plane at A, and on 
the vertical plane in the line A’A” perpendicular to the 
ground line, and GT'K’H' is the vertical projection of the 
frustum. Producing the line GY till it intersects A”A’, 
gives A’, the vertical projection of the vertex of the 
cone of which the frustum is a part. 

To find the shadow on the inner surface . 

If two tangent planes of rays be drawn to the cone, 
the points in which the elements of contact meet the 
upper circle, are the points where the shadow on the 
inner surface begins. 

Through (A, A’), the vertex of the cone, let a rav of 


SHADES AND SHADOWS. aay 


hight be drawn. Such ray pierces the plane of the upper 
base of the frustum in the point (B, B’). Through the 
point (B, B’), suppose two lines to be drawn tangent to 
_the upper circle of the frustum. These tangents are the 
traces, on the plane of the upper base, of two planes of 
rays drawn tangent to the cone. ‘The lines BC, BD, 
drawn tangent to the circle GDHL, are the horizontal 
projections of these traces; and (D, D’) (C, C’) are the 
points at which the shadow on the inner surface 
begins. 

Let the surface of the cone be now intersected by 
planes of rays passing through the vertex. Each se- 
cant plane so drawn will intersect the surface in two 
elements; the element towards the source of light 
will cast the shadow, and the other will receive it. 
Each plane will also contain the ray of light passing 
through the vertex of the cone, and consequently every 
trace on the plane of the upper base will pass through 
the point (B, Bb’). | 

Draw any line, as BaL, to represent the trace of such 
a plane. The horizontal projections of the elements in 
which it intersects the surface of the cone, are AL 
and Aa. Projecting L and a into the upper base at L’ 
and a’, and joining these points with A’, the vertical 
projection of the vertex of the cone, gives the vertical 
projections of these elements. 

Through (a, a’) the upper extremity of the element 
towards the source of light, conceive a ray of light to 
be drawn; the point (4, 4’) in which it meets the ele- 
ment (AL, A’L’) is a point of the curve of shadow. 

If we suppose the surface of the frustum to be pro- 
duced below the plane EF’, the shadow that would fall 
on the part of the surface below this plane is easily 
found. The lowest point is determined by passing a 


38 TREATISE ON 


plane of rays through the axis of the cone. The trace 
of this plane, on the upper base of the frustum, is hori- 
zontally projected in the line BAd. ‘The line A’‘d’ is the 
vertical projection of the element which receives the 
shadow, and (p, p’) is the point of shadow cast by (A, /’). 

By passing a plane of rays through the element (AH, 
A‘H’) we shall find the point at which the vertical projec- 
tion of the shadow is tangent to the element A'H’. The 
horizontal projection of the trace of this plane, on the 
upper base of the cone, is BkH. Projecting & into the 
vertical plane at /’, and drawing £'l parallel to the ver- 
tical projection of the light, gives 7 for the point of tan- 
gency. ‘The horizontal projection of the point is found 
by projecting / into the horizontal projection of the ele- 
ment at /. Having found as many points of the curve 
as are necessary to describe it accurately, let its pro- 
jections be made as in the figure. 

It is to be observed, that the shadow on the inner 
surface of the frustum intersects the plane of the lower 
base at the points (c¢,¢’), (f,/”); and that the curve of 
shadow below this plane is found on the supposition that 
the plane does not intercept the light, and that the frus- 
tum is produced below it. 

If the plane of the lower base be supposed to inter- 
cept the light, there will be no shadow on the surface 
of the cone below it; for the plane itself will receive 
the shadow cast by the upper circle. 

Through the centre (A, A”) of the upper circle, con- 
ceive a ray of light to be drawn. It pierces the plane 
of the lower base at (I, F’). A circle described with F 
as a centre, and radius equal to the radius of the upper 
base, is the horizontal projection of the shadow cast by 
the upper circle on the plane EF’. The arc eqf falls 
within the circle of the lower base, and therefore is the 


SHADES AND SHADOWS. 39 


line of shadow, within the surface, on the plane of that 
base. The circle described with the centre F will pass 
through ¢ and f, the horizontal projections of the points 
in which the curve of shadow intersects the plane E’F’. 

The shadow cast by the frustum on the plane of the 
lower base is limited by the traces of the two tangent 
planes of rays that determine the elements of shade. 
These planes contain the ray of light drawn through 
the vertex; hence their traces pass through the point 
(E, E’) in which this ray pierces the plane E’'F’; and En, 
Em drawn through E and tangent to the circle If Ke are 
their horizontal projections. 

The shadows cast by the elements of shade on the 
plane EE’ begin at the points in which they pierce it, 
and terminate at the points m and n, where the shadows 
are tangent to the shadow cast by the upper circle. The 
points m and x may be determined by drawing horizon- 
tal projections of rays through the points C and D, and 
noting their intersections with the shadow cast by the 
upper circle. 

That part of the shadow on the plane E’F’ which is 
under the surface of the cone is not seen in horizontal 
projection. 

For the purpose of showing the appearance of the 
shadow on the inner surface, we will suppose that part 
of the frustum which is in front of the vertical plane GAH 
to be removed, and then represent in vertical projection 
the remaining semi-frustum, as it appears to the eye. 

That part of the surface which lies between the ele- 
ment (AG, A’G’) and the curve of shadow, being in the 
shadow, is darkened in vertical projection. _ 


40 | TREATISE ON 


PROBLEM IX. 


Having given a sphere in space, and the direction of the 
light, ut ws required to find the curve of shade, and the shadows 
cast on the planes of projection. 


; 34. Let the centre of the sphere be taken at equal dis- 
. tances from the planes of projection; let A (PI. 5) be its 
horizontal and A’ its vertical projection; and suppose 
the projections of the light to make equal angles with 
the ground line. 

Suppose the sphere to be circumscribed by a tangent 
cylinder of rays. ‘The axis of the cylinder is the ray of 
light passing through the centre of the sphere; the 
curve of contact is a great circle whose plane is perpeéen- 
dicular to the axis of the cylinder, that 1s, to the direc- 
tion of the light in space. ‘This cirele of contact is the 
curve of shade. 

Since the axis of the tangent cylinder of rays 1s 
oblique to both planes of projection, the plane of the 
circle of shade which is perpendicular to it, is also 
oblique to both the planes of projection, and conse- 
quently its projections on these planes will be ellipses. 
(Des. Geom. 180.) 

To find the projection of the circle of shade on the 
horizontal! plane. 

The horizontal projection of that diameter of the cir 
cle of shade which in space is parallel to the horizontal 
plane, is the transverse axis of the ellipse into which the 
circle of shade is projected. The projection of that 
diameter which makes the greatest angle with the hori- 
zontal plane, or which is perpendicular to the parallel 
diameter, is the conjugate axis.—(Des. Geom. 180.) 


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SHADES AND SHADOWS. A} 


The ray of light through the centre of the sphere 
being perpendicular to the plane of shade, is also per- 
pendicular to that diameter of the circle of shade which 
is parallel to the horizontal plane, and therefore the pro- 
jections of this diameter and the ray of light,on the hori- 
zontal plane, are at right angles to each other.. Hence 
the diameter CAD, drawn through the centre A, and pet- 
pendicular to AB, the horizontal projection of the ray 
through the centre, is the transverse axis of the ellipse 
into which the circle of shade is projected. 

Through the centre of the sphere suppose a plane 
of rays to be drawn perpendicular to the horizontal 
plane. Its horizontal trace is EAB. This plane inter- 


seets the surface of the sphere in a great circle, the _ . 


cylinder of rays in two elements tangent to this circle, 
and the plane of the circle of shade in a diameter 
passing through the pots of contact. The projection 
of this diameter is the conjugate axis of the ellipse. 

To find the conjugate axis: 

Let this plane be revolved about a vertical axis pass- 
ing through the point A, till it becomes parallel to the 
vertical plane of projection. Inthe revolution, the point 
5 describes in the horizontal plane,the arc BD’, and the 
point (A, A’) being in the axis, remains fixed. Project- 
ing B’ into the ground line, and drawing A’B", we have 
the vertical projection of the revolved ray. But the 
great circle cut out of the sphere 1s,.when revolved par- 
allel to the vertical plane, projected into the circle rep- 
resenting the vertical projection of the sphere, and the 
elements cut from the cylinder of rays are projected 
into lines parallel to A’B”. 

Drawing therefore the tangents IH’ and G’b’ parallel 
to A’B", we determine, in the revolved position of the 
plane of rays, the extremities of that diameter of the 
circle of shade whose projection is the conjugate axis. 


a 


42 TREATISE ON 


When the plane of rays is parallel to the vertical plane, 
the point I” is horizontally projected at F, and the point 
G' at G. In the counter-revolution the points (1, F’) 
(G, G’) describe the horizontal arcs (F/, Ff), (Gg, 
G'g'), and gf becomes the horizontal projection of the 
diameter, or conjugate axis of the ellipse. Having 
found the two axes of the ellipse, let it be described. 

It is plain that (f,f’) is the highest point of the curve 
of shade, and (g, g’) the lowest; therefore the tangents 
to the curve of shade at these poits are horizontal, 
and their vertical projections parallel to the ground line. 

Since the points which are horizontally Praisetell at 
C and D are contained in the horizontal plane through 
the centre of the sphere, they will be vertically projected 
in its trace N’B”, at the points C’ and D’. 

The points which are horizontally projected at the 
intersections of the ellipse CgDf, with the line NB’, are 
vertically projected in the circumference of the circle 
representing the vertical projection of the sphere. We 
have thus found six poimts of the ellipse into which the 
circle of shade is projected on the vertical plane. And 
since the tangents passing through the highest and the 
lowest points f’ and g’ are parallel to the ground line, it 
follows that the diameter passing through these points is 
conjugate with the diameter D’C’; therefore let the 
curve be described (29). 

The axes of the ellipse into which the circle of shade 
is projected on the vertical plane can, however, be found 
by a construction similar to that used for the horizontal 
projection. 

Through A’ draw a diameter perpendicular to A’B, 
the vertical projection of the ray. This will be the 
transverse axis of the ellipse. Through the centre of 
the sphere conceive a plane of rays to be drawn per- 


SHADES AND SHADOWS. 43 


pendicular to the vertical plane—A’B is its vertical trace. 
Let this plane be revolved about an axis passing through 
the centre of the sphere and perpendicular to the ver- 
tical plane, until it becomes parallel to the horizontal 
plane. The point B describes in the vertical plane the 
arc BB”; and since (A, A’) remains fixed, AB” is the 
horizontal projection of the revolved ray passing through 
the centre of the sphere. The lines cH, hd’ drawn 
parallel to the revolved ray and tangent to the circle 
NDC, determine (c, c’), (h, h’), the extremities of the 
diameter, in its revolved position, whose vertical projec- 
tion is the conjugate axis of the ellipse. In the counter 
revolution, the points (c, ¢’) and (h, h’) describe the arcs 
(cd, cd’) and (hk, Wk’): and kd’ is the conjugate axis of 
the ellipse into which the circle of shade is projected on 
the vertical plane. In horizontal projection, we see only 
that part of the shade which lies above the horizontal 
plane N’B’”’; the part of the curve of shade lying below 
this plane is dotted. In vertical projection, we see that 
part of the shade which lies in front of the plane NB’. 

To find the shadow on the horizontal plane: 

The curve of shadow on the horizontal plane is the 
mtersection of the horizontal plane with the surface of 
the cylinder of rays tangent to the sphere. This curve 
will be an ellipse, unless the circle of shade be parallel 
to the horizontal plane, or the section a sub-contrary 
one. When the plane of rays AB was revolved par- 
allel to the vertical plane, we drew F’H’ and G0’ par- 
allel to the revolved ray A’B”. The tangent (G’d’, Gd) 
pierces the horizontal plane at 6. In the counter 
revolution, the point 6 describes in the horizontal plane 
the arc 6 6’; then 6’’B is the shadow cast by the radius 
of the circle of shade which passes through the lowest 
point—hence it is the semi-transverse axis of the ellipse 


44 TREATISE ON 


of shadow. The conjugate axis of this ellipse is the 
shadow cast by the horizontal diameter of the circle of 
shade. It passes through B,is perpendicular to 6B, and 
equalto CD. ‘The shadow cast on the vertical plane is 
found in a manner entirely similar. 


OF BRILLIANT POINTS. } 

35. When a ray of light falls upon a surface which 
turns it from its course and gives it another direction, 
the ray is said to be reflected. ‘The ray, as it falls upon 
the surface, is called the incident ray, and after it leaves 
the surface, the reflected ray. The poimt at which the 
reflection takes place is called the point of incidence. 
it is ascertained by experiment, 

1°. That the plane of the incident and reflected rays is 
always normal to the surface at the point of incidence. 

2°, That at the point of mcidence, the incident and 
reflected rays make equal angles with the tangent plane 
or normal line to the surface. 

If therefore, we suppose a single luminous point, and 
the light emanating from it to fall upon any surface and 
to be reflected to the eye, the poimt at which the 
reflection takes place is called the brillant point. 
The brilliant point of a surface is, then, the point at 
which a ray of light and a line drawn to the eye make 
equal angles with the tangent plane or normal line—the 
plane of the two Imes being normal to the surface. 

36. The rays of light being parallel, and the place of 
the eye at an infinite distance, the brilliant point of any 
surface is thus found: 

Through any point in space draw a ray of light and 
a line to the eye, and bisect the angle included between 
them. Thendraw a tangent plane to the surface and per- 
pendicular to the bisecting line—the point of contact is 


SHADES AND SHADOWS. 45 


the brilliant point. For, let AC (Fig.n) be the line drawn 
to the eye, AB the ray of light, AD the bisecting line, 
and P the point of contact of a plane passed perpendi- 
cular to AD, and tangent to the surface. Through P 
let there be drawn a ray of light PG,which will be paral- 
lel to AB, the line PE to the eye,which will be parallel to 
AC, and PF the normal line to the surface at the point P. 

Since the normal PF is perpendicular to the tangent 
plane, it is parallel to the bisecting line AD, for AD is 
perpendicular to the tangent plane by construction. But 
the line AD is in the plane of the lmes BA, AC, and 
bisects the angle BAC: therefore the parallel PF lies in 
the plane of the lines GP, PE, and bisects the angle 
‘GPE. Hence P is the brilhant point. 

36. To apply these principles in finding the brilliant 
point on the surface of a sphere. 

Suppose the eye to be in a line perpendicular to the 
vertical plane, and at an infinite distance from it. 

Through (A, A’), the centre of the sphere, suppose a 
ray of light to be drawn, and also a line to theeye. For 
the purpose of bisecting the angle included between these 
‘ lines, let their plane be revolved about (AP, A’) the line 
drawn to the eye, until it becomes parallel to the horizon- 
tal plane. Any point of the ray, as (E, E’), will describe 
an arc (Ke, He’); and Ae and AP are the horizontal pro- 
jections of the lines when revolved parallel to the horizon- 
tal plane. Bisect then, the angle PAe by the line Aq: and 
from any point of the axis, as P, draw the line Pge. After 
the counter revolution, the point e is horizontally pro- 
jected at E; and since P remains fixed, Pge is horizon- 
tally projected in PE; the point q of the bisecting line, 
is projected at q’, and Aq’ is the horizontal projection of 
the bisecting line. Its vertical projection is A’E’. 

The plane drawn perpendicular to this bisecting line, 


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46 TREATISE ON. 


and tangent to the sphere, touches the surface at the 
point wher’ the bisecting line pierces it—that is, at 
(a’, a”). Consequently, (a’, a’) is the brilliant point. A 
similar construction would determine the brilliant point, 
if the eye were taken in a line perpendicular to the hori- 
zontal plane. 


% 


PROBLEM X. 


Having given an ellipsoid in space, and the direction of the 
hight, vt 1s required to find the curve of shade, and the shadow 


cast on the horizontal plane. 


37. Let the horizontal plane be taken perpendicular to 
the axis of the surface. Let A (PI. 6. Fig. 1) be the 


horizontal projection of the axis, and A’B its vertical 


projection. Let us suppose the ellipsoid to be circum- 
scribed by a tangent cylinder of rays. ‘The axis of the 
cylinder is a ray of light passing through the centre 
of the ellipsoid and piercing the horizontal plane at C. 
Through this axis let a plane of rays be passed perpen- 
dicular to the horizontal plane—ACD is its trace. 

Since this plane is a plane of rays, and divides the ellip- 
soid into.two equal and symmetrical parts, the parts of 
the curve of shade lying on either side are equal and 


~ symmetrical. Hence, both parts are projected on the 


meridian plane AC, into the same right line. But, since 
the contact of the ellipsoid and surface of the cylinder 
is an ellipse whose plane passes through the centre of 
the ellipsoid, the curve of shade is projected into a right 
line passing through the centre, which line is the inter 
section of the plane of the curve of shade with the 


surface of the ellipsoid in a meridian curve, and the sur- 


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SHADES AND SHADOWS. 47 


face of the cylinder of rays in two elements that are 
tangent to it. The points of tangency are the highest 
and lowest points of the curve of shade. The line in 
which the plane of shade intersects the meridian plane 
AC, also passes through these points. 

Let the ellipsoid be now projected on a vertical plane, 
parallel to the meridian plane ACD. Assuming E’D’ 
for the new ground line, the centre of the ellipsoid will 
be projected at I’, in the perpendicular AF to the ground 
line E’D’, and at a distance from it, equal to the distance 
of the centre of the ellipsoid from the horizontal plane. 
The ray of light through the centre will be vertically 
projectedin FC’. The ellipse described about the cen- 
tre F, and equal to a meridian curve of the surface, 
represents the vertical projection of the ellipsoid. 

Having described this ellipse, draw the two tangents 
d’D’ and e’E’ parallel to FC’, the vertical projection of the 
ray. The points of contact d”,¢’ are the vertical pro- 
jections of the highest and lowest points of the curve 
of shade; and d”Fe’ is the vertical projection of the 
curve of shade, since the plane of shade is perpendicu- 
lar to the new vertical plane. The horizontal projec- 
tions of the highest and lowest points are d and e. 

Let us suppose a system of horizontal planes to inter- 
sect the ellipsoid between the highest and lowest points 
of the curve of shade. Each of such secant planes, 
being perpendicular to the axis, will intersect the sur- 
face of the ellipsoid in a horizontal circle, and the plane 
of shade in a horizontal le perpendicular to the meri- 
dian plane ACD, and consequently, to the new vertical 


plane. 
Let f’n be the vertical trace, on the new vertical plane, , 


of one of the secant planes. The line f'n is the vertical 
projection of the circle in which the plane intersects the 


48 TREATISE ON 


surface of the ellipsoid, and h” of the line in which it 
intersects the plane of shade. Projecting.the circle 
and line on the horizontal plane, the points fh and &, in 
which they intersect, are two points in the horizontal 
projection of the curve of shade. In a similar manne 
any number of points may be determined. The hori 
zonta] plane through the centre of the ellipsoid, inter 
sects the plane of shade in a line whose horizontal pro- 
jection is st, and this line is the transverse axis of the 
ellipse into which the curve of shade is projected; and 
ed, the projection of the diameter joining the highest 
and lowest points, is the conjugate axis. 

The projection of the curve of shade on the primitive 
vertical plane, is found by projecting the horizontal cir- 
cles, and drawing perpendiculars to the ground line 
through points of t the horizontal projection of the curve. 
Thus, projecting the horizontal circles passing through 
the highest and lowest pomts, and drawing perpendicu- 
lars to the ground line from d and e, determines d’ and 
eé, the vertical projections of the highest and lowest 
points. The vertical projection of the circle f’h'n, is Wk’, 
and h’,k’ are the vertical projections of the points of 
shade that are horizontally projected ath and k. The 
diameters d’e and /s’ are conjugate. 

The part of the surface which is in the shade, and in 
front of the meridian plane gAp, is shaded in vertical 
projection. ‘The part in the shade and above the 
horizontal plane passing through the centre of the ellip- 
soid, is shaded in horizontal projection. 

To find the shadow on the horizontal plane: 

If we suppose the lines in which the horizontal 
secant planes intersect the plane of shade, to cast 
shadows on the horizontal plane, the shadows cast 
will be parallel to the lines themselves. The line in 


SHADES AND SHADOWS. AY 


which the horizontal plane /’n intersects the plane of 


Mw 


shade, casts the shadowh’"/”, and the points h” and ”” in 
which it is intersected by hh” and £k”, drawn parallel to 
the horizontal projection of the light, are pomts in the 
shadow on the horizontal plane. The highest and 
lowest points cast their shadows at D and E. 

If through the highest and lowest points of the curve 
of shade, two planes be drawn tangent to the cylinder 
of rays circumscribing the ellipsoid, their traces on the 
horizontal plane will be tangent to the ellipse of shadow; 
but since the planes are perpendicular to the vertical plane 
EAD, their traces will be perpendicular to ED, at the 
points E and D. Hence, ED is an axis of the ellipse of 
shadow. The other axis passes through the middle 
point C, and is the shadow cast by the diameter (st, s‘7’). 


PROBLEM XI. 


Having given the plan and elevation of a niche, and the 
direction of the light, it is required to find the shadow which 
the niche casts upon tiself. 


38. A niche is a recess in the wall of a building. It 
is generally composed of a semi-cylinder, and the quad- 
rant of a sphere having the same radius as the base of 
the cylinder. ‘The quadrant of the sphere rests on the 
upper base of the semi-cylinder, forming the upper part 
of the niche. ‘The quadrant and semi-cylinder are tan- 
gent to each other in the semicircle which separates the 
cylindrical from the spherical surface. 

Let AB (Pl. 7. Fig. 1) be the intersection of the face 
of a vertical wall with a horizontal plane, and the semi- 
circle ACB the plan of the niche. The rectangle A’B’ 
is the elevation of the cylindrical part, and the semi- 

D 


50 TREATISE ON © 


circle A”F’B” of the spherical part of the niche—A’B’ 
is the vertical projection of the semicircle that separates 
the cylindrical from the spherical surface. 

The lines of the niche which cast lines of shadow on - 
the surface, are, 

1°. ‘The element (A, A’A”). 

2°. The semicircle A’ FB”. . 

A part of the shadow cast falls on the base of the 
niche, a part on the cylindrical, and a part on the spheri 
cal surface. 

To find the shadow on the base and cylindrical surface. 

Through the element (A, A’A”) conceive a plane of 
rays to be passed. Its horizontal trace AC is parallel to 
the horizontal projection of the rays of light, and the 
plane intersects the cylindrical surface in a second ele- 
ment (C, C’a), opposite the source of light, and on this 
element the shadow falls. The line A’a, drawn through 
A” parallel to the vertical projection of the rays of light, 
limits the shadow on the element: and drawing through 
C’ the line C’d, parallel to the vertical projection of the 
ray of light, determines the point 6, which casts a shadow 
atC. ‘The point C is common both to the cylindrical 
surface and to the base of the niche. Hence, the part 
A’b of the element (A, A’A”) casts the shadow AC on 
the base of the niche, and the part 6A”, the shadow. 
C’a on the cylindrical surface. Above the point a, the 
shadow on the cylindrical surface is cast by the semi- 
circle (AB, A’F’B”). 

' Through any point of the semicircle, as (I, F’), con- 
ceive a vertical plane of rays to be passed. Its hori- | 
zontal trace Ff is parallel to the horizontal projection 
of the rays of light, and the plane intersects the surface 
of the cylinder in the element (f, ff’). The point f’, 
in which the vertical projection of the ray drawn through 


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SHADES AND SHADOWS. 51 


F’ meets (f'f”), is the vertical projection of the shadow 
cast by the point (1*°,F”). In a similar manner, we may 
determine other points of shadow on the cylindrical 
surface. 

Above the line A”B” the shadow will fall on the spheri- 
cal surface. Before finding this shadow we will demon- 
strate, that when a cylinder intersects a sphere, the 
curve in which it enters on the one side, is equal to the 
curve in which it leaves the sphere on the other. 

For, the parts of the elements of the cylinder inter- 
cepted between the points at which they enter, and the 
points at which they leave the sphere, are parallel chords 
of the sphere. Conceive a plane to be passed through 
the middle point of one of these chords and perpendicular 
to it. Such plane passes through the centre of the 
sphere and bisects all the other parallel chords. Hence, 
the curves in which the cylinder enters and leaves the 
sphere, are symmetrical with respect to this plane, and 


are consequently, equal. Therefore, if one of themis 


a circle, the other will be an equal circle; and hence,  * 


when the surface of a cylinder intersects that of a sphere 
in a great circle, it will at the same time intersect it in 
a second great circle, and the planes of these circles 


will intersect each other in a diameter of the sphere. 


The two elements of the cylinder, passing through the 
extremities of this common diameter, are perpendicular 
to it, since they are tangent to the sphere. 

Let us suppose an entire hemisphere to be described 
on the diameter (AB, A’B”), having the plane of its 


great circle vertical. Through this circle suppose a | 


cylinder of rays to be passed. Half the surface of this 
cylinder intersects the surface of the hemisphere in a 
semicircle, which is the shadow that the semicircle 
(AB, A’F’B") would cast on the hemisphere... . 


2 a , 


52 TREATISE ON 


The plane of the circle casting the shadow, and the 
plane of the circle of shadow, intersect each other in a 
line passing through the centre of the hemisphere and 
perpendicular to the elements of the cylinder of rays, 
that is, perpendicular to the direction of the light. But, 
since this diameter is a line of the vertical plane AB, it 
is parallel to the vertical plane of projection—hence, its 
vertical projection is at right angles to the vertical pro- 
jection of the rays of light (Des. Geom. 49). ‘There- 
fore, if through D’, we draw the line D//'D” perpendicular 
to the projection of the ray, we determine Ih’, the ver- 
tical projection of the diameter in which the plane of the 
circle casting the shadow intersects the plane of the 
circle of shadow. The semicircle IA’) casts the 
semicircle of shadow on the hemisphere, the vertical 
projection of which shadow is an ellipse, whose trans- 
verse axis is Ih’ (Des. Geom. 180). 

Through (D, D’) let a plane of rays be passed per- 
pendicular to the face of the niche, and let us suppose 
for a moment the vertical plane of projection to be 
moved forward to coincide with this face. The point 
D’ will then be directly over D, and n'D'p will be the trace 
of the plane of rays. 

Let us now assume an auxiliary plane of projection, 
parallel to the plane whose trace is np, and at a given 
distance from it. Draw n’y’ parallel to n'p, to represent 
the trace of the auxiliary plane, and let it be borne in 
mind that this trace, as well as n’p, is in the plane of the 
face of the niche. The semicircle in which the plane 
of rays n'p intersects the hemisphere, is projected on 
the auxiliary plane in the semicircle n’hp’. The pro- 
jections of the ray of light passing through the point 
(n,n), are nk and nD’. This ray is projected on the 
auxiliary plane by projecting 7’ into the trace at n”, and 


SHADES AND SHADOWS. 53 


projecting any other point of the ray, as (4, D’) (Des. 
Geom. 14): this is done by laying off Dk’ equal to Dk 
The line n"&l is the projection of the ray on the aux 
iliary plane, and the point J, in which it meets the semi 
circle n/p’, is the projection of a point of the curve of 
shadow. But smce D‘h’, a line in the plane of the cir- 
cle of shadow, is perpendicular to the plane of rays 
whose trace is np, and consequently to the auxiliary 
plane, it follows, that the plane of shadow 1s _ per- 
pendicular to these planes; hence, the projection of the 
shadow on either of them is a right line (Des. Geom. 
82). But D” is the projection of h', a point of the curve 
of shadow, and a second point is projected at J; hence, 
Dl is the projection of the shadow on the auxiliary 
plane. This line is also the projection of the intersec- 
tion of the plane of rays, whose trace is n'p, with the 
plane of the circle of shadow. 

In space, therefore, the line whose projection on the 
auxiliary plane is D’/, is perpendicular to the diameter 
Ih’: hence, D7?’ its projection on the vertical plane, is 
the semi-conjugate axis of the ellipse into which the 
circle of shadow is projected (Des. Geom. 180). Let 
the ellipse be described. [rom h’ to the point q’, where 
the ellipse intersects at A"B", the shadow falls on the sur- 
face of the sphere. From q' the shadow falls on the 
cylindrical surface. The arc /’E casts the shadow on 
the spherical, and the are A”E the shadow on the cylin- \ 
drical surface. 

Had the quadrant of the sphere below the plane A” B’ 
been permitted to intercept the light, the shadow af’q, 
instead of being on the cylinder, would have been the 
shadow q/l'I on the surface of the sphere. We have also 
dotted the line of shadow ae'vg’, which the front circle 


54 TREATISE ON 


of the hemisphere would cast upon the vertical cylinder, 
if the sphere did not intercept the rays of light. 

The part of the surface of the niche lying between 
the lines A’A” Ef’ and the line of shadow, is shaded in 
vertical projection. 

It is not necessary to find the semi-conjugate axis of 
the ellipse into which the circle of shadow is projected. 
We may find points of the curve, and describe it through 
them. 

Let z's’ be the vertical trace of a plane of rays per- 
pendicular to the face of the niche. This plane inter- 
sects the hemisphere ina semicircle which is projected 
on the auxiliary plane in the semicircle described with 
the radius D’?”.. Draw2"s” parallel to nl. The point s”, _ 
where it meets the semi-circle and the line DJ, is the pro- 
jection on the auxiliary plane of a point of the circle of 
shadow. This point is projected on the vertical plane 
of projection at s’, and on the horizontal plane at s, by 
making ms equal to m’s”. In a similar manner other 
points of the curve may be found. 

The point A’ is projected on the horizontal plane at h, 
and the point g at qg: therefore, hsq is the horizontal pro- 
jection of the shadow which falls on the spherical part 
of the niche. The shadow on the cylindrical part is 
horizontally projected in the are Ceg. 

The point (q, 9), at which the shadow passes from 
the cylindrical to the spherical part of the niche, can be 
found by a direct construction. 

The point in space, is the one in which the intersec- 
tion of the upper base of the cylinder with the plane of 
shadow, meets the circle of the upper base. But h'1 is 
the trace of the plane of shadow on the face of the niche, 
and (s, s') is a pot of this plane. Therefore, we have 
one trace of an oblique plane, and a point of the plane, 


SHADES AND SHADOWS. 55 


to find its other trace (Des. Geom. 43). Draw su 
parallel to h’D’; its horizontal projection is su, and (u, wv’) 
is the point in which it pierces the horizontal plane A’ B": 
therefore, (uw, u’) is a point in the trace of the plane of 
shadow. But the trace passes through (D, D’); hence 
Dug is the horizontal projection of the required trace. 
But q' is the vertical projection of g; hence (q, q’) 1s 
the point at which the shadow passes from the cylin- 
drical to the spherical surface. 


PROBLEM XI. 


To find the curve of shade on the surface of a torus. 


39. Let abcg (PI. 7, Fig. 2) be a rectangle, having the 
semicircles aA’é and cB'g described on its vertical sides. 

If the figure aA’bcB'g be revolved about a vertical axis 
(C, C”O), it will generate a solid, called a torus. This 
solid is used in some of the orders of architecture in 
forming the bases and capitals of the columns. 

Before finding the curve of shade, we will demonstrate, 
that if a surface of revolution be intersected by a me- 
ridian plane, and a line be drawn tangent to the meridian 
curve and parallel to the projection of the rays of light 
on the meridian plane, the point of contact will be a 
point of the curve of shade. 

For, every plane of rays, tangent to a surface, touches 
it ina point of the curve of shade. But since the surface 
is one of revolution, such plane is perpendicular to,the 
meridian plane passing through the point of contact (Des. 
Geom.105). ‘Therefore, its trace will not only be tangent 
to the meridian curve, but also parallel to the projection 
of the rays of light, since the rays are projected by per- 


ned 


56 TREATISE ON 


pendicular planes. Hence, the tangent to a meridian curve, 
drawn parallel to the projection of the rays of light on ets plane, 
determines a point of shade. 

Let the circle ATBH represent the horizontal pro- 
jection of the surface, ¢B’cbA’a its vertical projection, 
and (A, A’) the ray of light. Through any point of the 
axis of the surface, as (C, ©’), suppose a ray of light to 
be drawn, CD, C’D’ are its projections. ‘Through this 
ray let a meridian plane be passed, and then revolved 
about the axis of the surface until it becomes parallel to 
the vertical plane of projection. The point (C, C’) being 
in the axis, remains fixed, and any point of the ray, as 
(D, D’), describes a horizontal arc (Dd, D'd’); and the 
vertical projection of the ray, from its revolved position, 
is C’d’. The section of the surface, when revolved 
parallel to the vertical plane, is vertically projected in 
the meridian line gB’cbA’a. 

If now, two lines be drawn tangent to the semicircles 
and parallel to C’d’, the revolved ray, they will determine 
the points (£, £’) and (¢, 2’): these are the highest and low- 
est points of the curve of shade, in their revolved position. 
In the counter revolution of the meridian plane, they 
describe horizontal arcs, and when the counter revolu- 
tion is completed, are horizontally projected at h and J, 
and vertically at A’ andl’. 

The lines drawn parallel to C’D’, the vertical projec- 
tion of the ray, and tangent to the semicircles, determine 
the points (q,9') and(p,p') at which the curve of shade ’ 
intersects the meridian plane AB. 

The points (T, ¢’) and (H, 7”) of the curve of shade 
are found by passing two planes of rays tangent to the 
surface and perpendicular to the horizontal plane. The 
planes will touch the surface in the circumference of the 
circle whose vertical projection is AB’. The points of 


SHADES AND SHADOWS. 57 


tangency are also found in a meridian plane passed per- 
pendicular to the plane DCA. 

To find other points of the curve of shade: 

Draw any meridian plane, as nCm, and project the ray 
of light upon it. The point (C, C’) is its own projec- 
tion, and (D, D’) is projected at (F, fF’); therefore, (CF, 
C’F’) is the projection of the ray on the meridian plane 
nCm. Let the meridian plane be revolved till it becomes 
parallel to the vertical plane of projection. The line (CF, 
C’F’) will then be vertically projected inthe lineC/’. Draw 
two lines parallel to Cf’ and tangent to the semicircles; 
their points of contact (7, 7’) and (v, v’) are points of the 
curve of shade in their revolved position. In the counter 
revolution these points describe horizontal arcs, and 
when. it is completed are horizontally projected at n 
and m, and vertically atm and m. Any number of points 
of the curve of shade may be found in a similar manner. 
The curve HqlTphmis the horizontal,and ql'n (pm the 
vertical projection of the curve of shade. 


PROBLEM XUIL 


Having given a surface of revolution, conver towards the 
axis, and the direction of the light, it 1s required to find the 
shadow which the upper circle casts upon the surface, and also 
the brilliant point. 


40. Let DRN (PI. 8, Fig. 1) be the horizontal, and 
D'L the vertical projection of a cylinder or pedestal, on 
which the surface rests. 

Take two-thirds of the radius AD, and lay it off from 
A’ to v, on the vertical line A’B. Through v draw the 
horizontal line vu, which meets the vertical line D’u at 
#«. With was a centre, and radius wD’, describe the 


SOF . (PREATISE ON 

quadrant D’H’X’. Then ‘lay off vB equal to one-third 
of the radius A’D’, and draw Bl’ parallel to A’D’, and 
equal to two-thirds of it. Withy asa centre, and a radius 
equal to yX’ or yl’; describe the quadrant X’dl’. The 
two quadrants will have a common tangent line at the 
point X’, which will be vertical. Now, ifthe curve D'H’X’ 
be revolved about the vertical axis (A, A’B) it will gene- 
rate a surface of revolution, convex towards the axis, 
and concave outwards. The circle whose radius is X’v 
is called the circle of the gorge. 

Let us also suppose a SringeR Hayne the radius of its 
base equal to the radius of the upper circle of the sur- 
face, to be placed on the surface. 

It is required to find the curve of shadow which the 
upper circle (daQS, /'z’) casts upon the surface, under the 
supposition that the light is not intercepted by the sur- 
face. Admitting this supposition, it follows, that each 
point of the upper circle casting a shadow will, in gene- 
ral, cast two points of shadow; one where the ray 
through the point enters the surface, and the other 
where it leaves the surface. 

Through the upper circle of the surface suppose a 
cylinder of rays to be passed. The ray (AC, BC’) is the 
axis of this cylinder, and the curve in which it intersects 
the surface of revolution is the curve of shadow r equired. 

Through the axis of the cylinder of rays and the axis 
of the surface, suppose a meridian plane to be passed. 
Its horizontal trace is AC, and it cuts the upper circle 
of the surface in two points, one of which (a, a’) casts 
shadows on the meridian curves. Let this plane be 
revolved parallel to the vertical ,plane of projection. 
After it is so revolved, the ray of light will be vertically 
projected in Be’, the meridian curves in the curves repre- 
‘senting the vertical projection of the surface, and the 
point (a, a’) at 7’. o 


as 


ks 


E 
i Tl 


N 


— 
=n 
SSS 
————— 


| 
| 


| 


| WIN 


——— —- --__ eee? 


LE. Prudhomme St, 


pe 


ts 


SHADES AND SHADOWS. 59 


Through /’ draw /'éd parallel to the revolved ray Be’; 
the points 6 and d in which it intersects the meridian 
curves, are the highest and lowest points of the curve 
of shadow, in their revolved position. In the counter 
revolution the points 6 and d describe arcs of horizontal 
circles, and when the revolution is completed, are hori- 
zontally projected at and g, and vertically at h’ and ¢’. 
These points may also be found by drawing through », 
the point in which the revolved ray l’bd meets the axis, 
the line g’vh' parallel to BC’, the vertical projection of 
the ray, and noting its intersection with the horizontal 
lines dg’ and 6h’; and then projecting the points g’ and 
h’ into the horizontal plane in the line aAC. 

To find points of the curve between the highest and 
lowest points, we intersect by horizontal planes. Each 
horizontal plane will intersect the surface of revolution 


in a circle, and the surface of the cylinder of rays in a~ 


circle equal to the upper circle of the surface : the points 
in which these circles intersect are points of the curve 
of shadow. 

Let H'C’ be the trace of an auxiliary horizontal plane. 
This plane cuts the axis of the cylinder of rays in the 
point (C, C’). With C as a centre, and a radius equal 
to Bz’, describe the arc pm; this will be the horizontal 
projection of an arc of the circle in which the auxiliary 
horizontal plane intersects the surface of the cylinder 
of rays. Projecting H’ into the line DA, we have AH 
for the radius of the circle in which the same plane inter- 
sects the surface of revolution. Describing that circle, 
and noting the points in which it intersects the circle 
described witl the centre C, we find m and p, the hori- 
zontal projections of two points of the curve of shadow. 
These points are vertically projected in the vertical 
trace of the auxiliary plane, at m’ and »’. Similar con- 


ee ae 
“ ' 


60 TREATISE ON 


structions determine the points (2, 2’), (e, €), (0, 0’), and 
(n,n). ‘The points (2, 2’) and (e, e’) are im the circle of 
the gorge. 

That part of the curve is made full, in vertical pro- 
jection, which is in front of the meridian plane DAz; 
and the part of the surface which is above the curve 
of shadow is shaded. The elements of shade on the 
pedestal are (N, N’) and (R, R’); and on the upper 
cylinder (Q, Q’) and (8, 8’). 

41. To find the brilliant point : 

Suppose the eye to be situated in a perpendicular to 
the vertical plane, and at an infinite distance from it. 

Through any point of the axis, as (A, A’), suppose a 
ray of light to be drawn, and also a line to the eye, and 
the angle contained between them to be bisected as in 
Art. 36. The bisecting line is (AK, AK’). It is now 
required to pass a plane perpendicular to this line and 
tangent to the surface; the point of contact will be the 
brilliant point. 

If we suppose the tangent plane to be drawn, its trace 
on the meridian plane passing through the bisecting line 
(AK, A’K’) will be perpendicular to the bisecting line, 
and tangent to the meridian curve. Let this meridian 
plane be revolved parallel to the vertical plane of pro- 
jection. The bisecting line will then be vertically pro- 
jected in the line A’é’, and the meridian curve im the curve 
DHT’. Let H'l be drawn perpendicular to A’k’, and 
tangent to the curve D'H’; the point of contact (H, H’) 
is the revolved position of the brilliant point. Draw 
the normal H’V perpendicular to the tangent, or parallel 
to A‘k’. In the counter revolution, V being in the axis 
remains fixed, and H’ describes the arc of a horizontal 
circle. After the counter revolution, the bisecting line 
is vertically projected in A’K’, and the normal VH’ in 


SHADES AND SHADOWS. 61 


VP’, parallel to A’K’. Hence P’, where the horizontal 
line H'P’ intersects VP’, is the vertical projection of the 
brilliant point.. its horizontal projection is at P, in the 
horizontal trace of the meridian plane AK. 

The construction here given, is general for all sur- 
faces of revolution. If the eye is supposed to be in 
a line perpendicular to the horizontal plane, the brilliant 
point is easily found; for we bisect the angle as before, 
and draw a tangent plane perpendicular to the bisect- 
ing line. | 

It is plain that a second line can be drawn perpen- 
dicular to £'A’, produced on the other side of A’, which 
shall be tangent to the meridian curve 2d. A second 
tangent plane can therefore be drawn perpendicular to 
the bisecting line, and the point of contact will answer 
the mathematical conditions of a brillant point. ‘The 
point, however, will be on thas’ part of the surface which 
is not seen by the eye. 


PROBLEM XIV. 


Having given a surface of revolution and the direction of 
the hght, tt 1s required to find the curve of shade. 


42. Fig. 2, Pl. 8 represents the projections of a sur- 
face of revolution generated as in the last problem. 

Every ray of light that is tangent to the surface of 
revolution is an element of the tangent cylinder of rays 
which determines the curve of shade; therefore, every 
point at which a ray of light is tangent to the surface 
is a point of the curve of shade. 

Through the axis of the surface let a meridian plane 
of rays be drawn—PB is its horizontal trace. Let this 
plane be revolved until it becomes parallel to the verti- 
cal plane of projection. The ray of light through (A, A’) 


62 TREATISE ON 


will then be vertically projected in the line A’b’, and the 
meridian curves, in the curves which represent the ver- 
tical projection of the surface. Let the two tangents 
ad and ge be drawn parallel to the revolved ray A’b’; the 
points of contact (a, a’) and (e,¢) are the highest and 
lowest points of the curve of shade in their revolved 


position. After the counter revolution, these points are — 


horizontally projected at c and f, and vertically at ¢ and 
f. The lines gf’ and dé are parallel to A’B’ the verti- 
cal projection of the ray. 

To find other points of the curve of shade, we use 
auxiliary tangent surfaces. 

Draw any line, as E’Z’, between the highest and lowest 
points. At E’ or #’, draw a tangent, as EC, to the meridian 
curve. Conceive the meridian plane and the right-angled 
triangle E’CO to be revolved about the axis of the sur- 
face. The meridian curve generates the surface of revo- 
lution, and the line EC the surface of a right cone tan- 
gent to it ina circle whose vertical projection is E’k’, and 
horizontal projection Emn. If now, two tangent planes 
of rays be passed to this cone, they will be also tangent 
to the surface of revolution at two points in the circum- 
ference of the circle (Emn, E’k’): these points are points 
of the curve of shade. 

Through (A, C), the vertex of the cone, let a ray of 
light be drawn. This ray pierces the plane of the cone’s 
base at (D, D’). Through (D, D’) let two lines be drawn 
tangent to the base of the cone—these tangents are the 
traces of the two planes of rays that are tangent to the 
cone, and Dm and Dn are their horizontal projections. 
Projecting the points m and n into the vertical plane at 
m andn', we determine two points - (m,m’) and (n,7’) 
of the curve of shade. In a similar manner, other 
points of the curve of shade may be found. 


- 


SHADES AND SHADOWS. 63 


If the circle of contact (Emn, E’k’) be taken nearer 
the circle of the gorge, the vertex of the cone will be 
farther from the horizontal plane, and the elements will 
be nearer vertical. And when the circle of the gorge 
is assumed for the circle of contact, the auxiliary tan- 
gent surface will be a vertical cylinder, having a common 
axis with the surface of revolution. If two planes of 
rays be drawn tangent to this cylinder, the elements of 
contact will pierce the horizontal plane at z and /, the 
opposite extremities of a diameter perpendicular to the 
horizontal projection of the ray of light. This cylinder 
determines the two points (J, /’) and (2, z’) of the curve 
of shade. 

There is yet a third method of finding points of the 
curve of shade: it is by means of auxiliary tangent 
spheres. 

Assume any line, as p’r, for the vertical projection of 
the circle of contact of a sphere and the surface of 
revolution. At p’ draw a tangent to the meridian curve, 
and pg perpendicular to it. ‘The point g, where the 
perpendicular meets the axis, is the vertical projection 
of the centre of the sphere, and qp’ is its radius. Let 
us now suppose this sphere to be circumscribed by a 
tangent cylinder of rays. This cylinder will touch the 
sphere in a great circle, the plane of which will be per- 
pendicular to the direction of the ight. Therefore, the 
trace of the plane of the circle of contact of the sphere 
and cylinder of rays, on the meridian plane EAQ, or on 
the vertical plane of projection, will be perpendicular to 
the projection of the ray of light on either of these 
planes (Des. Geom. 49). Hence, qs’, drawn through 
gq perpendicular to A’B,, is the vertical projection of the 
line in which the plane of the circle of contact of the 
sphere and cylinder intersects the meridian plane EAS. 


64 TREATISE ON 


But the plane of this circle of contact intersects the 
plane of the circle of contact of the sphere and surface 
of revolution in a horizontal line perpendicular to the 
direction of the light. This horizontal line must pierce 
the meridian plane EAé in the trace qs’, and also in the 
trace p’r; therefore it pierces it at (s,s°). Hence, the 
line tsv, drawn through s perpendicular to the horizontal 
projection of the ray, is the horizontal projection of the 
intersection sought. Projecting p' mto the horizontal 
plane at p, and describing a circle with the radius Ap, 
we have the horizontal projection of the circle of con- 
tact of the sphere and surface of revolution. Projecting 
the points ¢ and v, in which the line ésy intersects this 
circle, into the vertical plane at ¢’ and v’, and we have the 
points (¢, ¢’) and (v, v') which are common to the surface 
of revolution, the tangent sphere, and the cylinder of rays. 

If through these points planes be drawn tangent to 
the cylinder of rays, they will be planes of rays, and 
tangent both to the sphere and surface of revolution. 
Hence, the points (/,¢’) and (v, v’) are points of the curve 
of shade. 

If we suppose the space within the surface of revo- 
lution to be unoccupied, a part of the curve of shade 
will cast a shadow that will fall onthe convex side of the 
surface. 

The ray of. light which determines (c, ¢’), the highest 
point of shade, being produced, will intersect the oppo- 
site meridian curve: the poit of intersection is a point of 
shadow. Thereare points of the curve of shade, on both 
sides of the meridian plane PAD, which also cast sha- 
dows on the surface. As we recede from the meridian 
plane PAD, on either side, the part of the ray inter- 
cepted between the point of shade and the correspond- 
ing point of shadow, continually diminishes, and finallv 


SHADES AND SHADOWS. 65 


becomes nothing, or the point of shadow unites with the 
point of shade casting it. 

At these two points, one on each side of the meridian 
plane PAD, the ray of light will be tangent to the curve 
of shade. For, when the point of shadow unites with 
the point of shade, they become consecutive points of 
the curve of shade; hence, the ray passing through 
them is tangent to it (Des. Geom. 65). But if the 
rays of light at these two points are tangent to the curve 
of shade, their projections will be tangent to the projec- 
tions of ‘the curve (Des. Geom. 90). The horizontal 
projection of the curve of shade being constructed, draw 
two tangents to it parallel to PAD, the horizontal pro- 
jection of the ray of light. The points of contact are X 
and y, and Xcy is the horizontal projection of that part 
of the curve of shade which casts a shadow on the 
interior of the surface. Projecting the pomts X and y 
into the vertica! projection of the curve, at X’ and 7’, and 
drawing lines parallel to the vertical projection of the 
rays, the lines so drawn will be tangent to the vertical 
projection of the curve of shade. 

Descending along the curve of shade, from the points 
(X, X’), (y, 7’), the rays of light touch the surface on the 
concave side, and the points of the curve of shade still 
cast shadows upon the surface. When we reach those 
points, one on each side of the meridian plane PAD, at 
which the point of the curve of shade unites with the 
point of the curve of shadow, the rays become tangent 
to the curve. Therefore, drawing two other tangents 
parallel to PD, their points of contact w and z are the 
horizontal projections of two points at which the rays 
of light are tangent to the curve of shade. At these 
points the curve of shade returns to the convex side of 
the surface. Projecting these points into the vertical 


66 TREATISE ON 


projection of the curve of shade, we find the points Ww 


and z’, through which, if the projections of raysbe drawn, .— 
they will be tangent to the vertical projection of the: 


~ curve of shade. 


That part of the curve of shade whose beaant 


projection is yeX lies on the convex side of the surface ; 
the part Xv/mz lies on the concave side of the surface ; 
the part zfw lies on the convex side, and the part wnity 
on the concave side. The curve is symmetrical with 
respect to the meridian plane PAD. : 
The part of the curve of shade which is in front, cof 

the meridian plane EAd is made full in vertical projec- 
tion, and the part of the surface lying above the curve 
of shade,and seen,is darkened. 


PROBLEM XV. 


Having given the position of a surface of revolution and 
the direction of the hght, tt ts required to find the line which 
separates the dark from the wluminated part of the surface, 
and the shadow which is cast on the horizontal plane of pro- 

jection. a 


- 43. Let (A, A’B), (PL 9) be the axis of the surface, 
and let the projections of the surface be made as In the 
figure. | 
Find the shadow which the upper horizontal circle 
CD casts upon the surface, as in Prob. 13, and Hien 
find the curve of shade, as in Prob. 14. | 

The highest and lowest points of the curves of mids 
and shadow, are in the meridian plane of rays EAB’: 

By considering the form of the meridian curves, it j 1s 
plain, that the highest point of the curve of shade is 
above mos highest point of the curve of shadow, and the 


™ « 


= + ra i? 
. & ; be 
r ie ype 
“ a x. ‘ * z 7 
, .. > : * . 
. = no : , ’ - - 
, ~ a . @ . - . 
“ + 
= Ss 2 - - ~ : 7 
~ a © v 
. . - e 4 ‘ i 7? . 
? ¢ by - 
* - . x 2 . 1 
‘ > . = . . 
‘ = *. oy * @ 
: : * ~ * - 
* ® - 
: ~ 
= P ~ + e 3 - oe et 
+ & w — ¥; 5, > 
~ 
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SHADES AND SHADOWS. 67 


lowest point of the curve of shade below the lowest point 
of the curve of shadow. ‘These curves will therefore 
intersect each other. This they do at the points (a, a’) 
and (6, 6’). Above these points the curve of shadow 
being below the curve of shade, and on the exterior sur- 
face, separates the dark from the illuminated part of the 
surface. 

Below these points, the curve of shade is below the 
curve of shadow, and separates the dark from the illu- 
minated part of the surface, until it returns to the con- 
vex side of the surface at the points (c, c’) and (d, d’). 

It has already been observed, in Prob. 14, that the 
parts of the curve of shade (fac, f'ac’) and (gbd, g‘b'd’), 
which are on the concave side of the surface, cast 
shadows upon it. These shadows begin at the points 
(c, c’) and (d, a’). 

To find these shadows, we will, in the first place, find 
the shadow which the entire curve of shade would cast 
on the horizontal plane, under the supposition that the 
surface offers no obstruction to the light. This is done 
by finding where rays of light, drawn through the several 
points of the curve of shade, pierce the horizontal plane. 

The upper part of the curve, which is on the convex 
side of the surface, casts the shadow /”h’g’—the lower 
part of the curve, on the convex side of the surface, 
casts the shadow c’l”d"—the parts of the curve on the 
concave side of the surface, cast the shadows c’t’f” and 

"g'g". Assume now any horizontal circle below the 
points (c, ¢’), (d, d’); the one, for example, whose ver- 
tical projection is &'mn, and whose horizontal projection 
is the circle described with the centre A and radius Ak © 
equal to mk’. The centre of this circle casts a shadow 
on the horizontal plane at m’—the circumference de- 
scribed with m’ as a centre, and radius mp, equal to mk’, — 


4% 


68 TREATISE ON 


will be the shadow cast on the horizontal plane by the 
circumference of the assumed circle. 

If through the points p and q, in which this shadow 
intersects the shadow cast by the curve of shade, rays 
of light be drawn, they will intersect, in space, the cir- 
cumference of the horizontal circle and the curve of 
shade. The pomts in which these rays mtersect the 
circumference of the horizontal circle, are points of 
shadow on the surface which are cast by the pomts in 
which the rays intersect the curve of shade. 

Therefore, draw the horizontal projections of rays 
through the points p and g, and the points p’ and q’, in 
which they intersect the circumference kp‘q, described 
with the centre A and radius As, are the horizontal 
projections of two points of the shadow on the sur- 
face. ‘These pomts are vertically projected at p” and 
gq’. By similar constructions, we may find any num- 
ber of points in the shadow which the curve of shade 
casts on the surface. 

If we take the lower horizontal circle FG, we shall 
find the points (7, /) and (x, x’) where the shadows on 
the surface terminate. The curve (cpt, cpt) and (dqz, 
d'qx°) may now be described. ‘The curve (cp’t, cp?) 
intersects the meridian plane HA at the point (s, s’). 

We have thus found the lines on the exterior of this 
surface, which separate the dark from the illuminated 
part. They are the curve of shadow until it intersects 
the curve of shade, then the curve of shade until it 
returns to the interior surface, and then the curve of 
shadow cast on the surface by the curve of shade. The 
light does not fall on that portion of the surface which 
is above and within the parts of these curves. 

Let us now find the shadow which the surface casts 
on the horizontal plane. 


SHADES AND SHADOWS. 69 


The elements of shade (I, I’) and (K, K’) cast lines of 
shadow onthe horizontal plane, which are found by 
drawing rays of light through their upper extremities. 
Then find the shadow cast by the circumference of the 
circle whose vertical projection is FG. This shadow 
intersects the shadow cast by the curve of shade in the 
points ¢” and 2”. ‘These points of shadow are cast by 
the points (¢, t’) and (a, 2’). » 

I’ind next, the shadow cast by the circumference of 
the upper circle of the surface. The centre of this cir- 
cle casts a shadow af B’.. With B’ as a centre, and a 
radius equal to BD, let the circumference of a circle be 
described—this circumference is the shadow sought. 

If through the points y and z, in which this shadow 
intersects the shadow cast by the curve of shade, rays 
of light be drawn, they will intersect, in space, both the 
curve of shade and the upper circle of the surface. | 
These points of shadow are therefore cast by the 
points (a, a’) and (6, 6’) in which the curves of shade and 
shadow intersect upon the surface. The shadows ¢’y 
and xz are cast by parts of the curve of shade. 

The elements of shade (P, P’), and (L, L’) of the 
upper cylinder, cast shadows which are tangent to the 
circumference described with the centre B’. The cir- 
cumference of the upper circle of the cylinder casts a 
shadow on the horizontal plane, which is also tangent 
to the shadows cast by the elements of shade. 


OF THE SHADES AND SHADOWS OF THE ROMAN DORIC 
COLUMN. 


44, This column is composed of three principal 
parts :—Ist. The base; 2d. the shaft; and 3d. the capital. 
‘That the figure may not be too complicated, we shall 


70 TREATISE ON 


first find the shades and shadows on the base and shaft 
of the column, and then use a separate figure to deter- 
mine those of the capital. 

45. The base of the column (PI. 10. Fig. 1) is com- 
posed of seven parts. 1. A rectangular prism, called 
a plinth, whose horizontal sections are squares: 2d. 
A solid of revolution, convex outward, called the lower 
torus: 3d.A cylindrical fillet: 4th. A solid of revolution, 
concave outward, called a scotia: 5th. A cylindrical 
fillet: 6th. The upper torus: 7th. A cylindrical fillet. 

The shaft of the column rises from the upper fillet. 
For a short distance it is concave outward, then it be- 
comes nearly cylindrical, and continues so to near its 
upper extremity, where it again becomes concave out- 
ward. ‘The drawing is made on the supposition that 
the shaft is cylindrical between the parts of it which are 
concave outward. 

46. The capital is composed of two distinct parts— 
the one a member whose horizontal sections are 
squares, having their centres in the axis of the column, 
the other a solid of revolution. The projections of 
the capital are shown in Fig. 2. The part of the capi- 
tal whose vertical projection is ww’o’s's, is the portion 
whose horizontal sections are squares—this part is 
called the abacus. ‘he part of the abacus between 
ry and nv is called the cyma-reversa, or talon. 

The part of the capital whose vertical projection is 
ll’mm is called the echinus. The part whose vertical 
projection is k&T7/ is called the upper fillet. The part 
whose vertical projection is 7k’% is called the cavetto. 
The part whose vertical projection is ge"/’h is called the 
neck. The part ff’ is called the astragal, or colarino. 
The part whose projection is dd’c’e is called the lower 


SHADES AND SHADOWS, 71 


fillet. ‘The entire column below the abacus and above 
the plinth is a solid of revolution, and the vertical projec- 
tions, in the two figures, are meridian sections of its 
surface. 


PROBLEM XVI. 


To find the shades and shadows on the shaft and base of the 


Roman Dorie column. 


47, A semi-column will illustrate all the cases. Let 
the semicircle AMB (PI. 10. Fig. 1) be the horizontal 
projection of the cylindrical part of the column, CND 
of the upper and middle fillets, EOF of the upper torus, 
GPH of the lower fillet, LQS of the lower torus, and 
the rectangle LR of the plinth. 

The space included between the semicircle GPH, the 
horizontal projection of the lower fillet, and a semi- 
circle described at equal distances from AMB and CND, 
iimits the horizontal projection of the scotia. Let the 
vertical projections be made as in the figure. 

Through the axis of the column, let a meridian plane 
of rays be drawn—VI isits horizontal trace. Through 
I draw la at right angles tolV. ‘The plane of rays, 
tangent to the cylindrical part of the shaft, touches it 
in an element, which pierces the horizontal plane at a; 
this element, which is the element of shade, can there- 
fore be drawn. 

The element of shade casts a shadow on the concave 
part of the shaft, beginning atthe point (a, a’). This 
shadow is in the vertical plane of rays passing through 
the element of shade, and is therefore horizontally pro- 
jected in the right line drawn through a perpendicular 
to la. The vertical projection of this shadow is found 


72 » °° * PREATISE ON 


by intersecting the shaft below a’ by horizontal planes. 
Each plane will intersect the shaft in a circle, which 
being projected on the horizontal plane, the circum- 
ference will intersect the tangent through a. ‘Then, pro- 
jecting the point of intersection into the vertical plane, 
we determine a point of the vertical projection of the 
shadow. 

The plane of rays tangent to the upper fillet, touches 
it in the element which is projected on the horizontal 
plane at 6. This element of shade, and the circum- 
ference of the upper circle of the fillet, cast shadows on 
the upper torus. The shadow of the element is found 
by a construction similar to that used in finding the 
shadow on the foot of the shaft. Points of the shadow 
cast by the circle are found by first finding the shadow 


_of the circle on a horizontal plane, and then finding the 


shadow which‘a horizontal section of the torus would 


cast on the same plane, and drawing rays through their 


points of intersection. ‘The shadow cast by the circum- 
ference of the upper circle of the fillet on the torus, is 


* made full, until it intersects at (A, h’) the curve of shade, 


determined as in Prob. 12. 

Passing the curve of shade on the upper torus, the 
construction for which is given in Prob. 12, we come 
next to the shadow which this curve casts upon the 
middle fillet. : 

To find this shadow, we intersect the torus and the 
fillet by a plane of rays cp’, perpendicular to the vertical 
plane of projection. Through the point ¢, in which 
the trace of this plane intersects the vertical projection 
of the curve of shade, let a horizontal plane be passed— 
this plane will intersect the torus in a horizontal circle, 
and let this circle be projected on the horizontal plane. 
The horizontal projection of ¢ is at c, and (¢,c’) is a 


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4 


SHADES AND SHADOWS. 73 


point of the curve of shade. The ray of light through 
this point pierces the fillet in the point (p, p’), which is 
a point of the curve of shadow. This shadow passes 
down. upon the fillet obliquely, until it intersects the 
lower circle of the fillet or upper circle of the scotia, 
at (X, X’). That part of the upper circle of the scotia, 
from h” to (X, X’), on which the light falls, casts a shadow 
on the scotia, which is found as in Prob. 13. 

The ray of light passing through the point (X, X’), in 
which the shadow on the fillet intersects the upper circle 
of the scotia, determines the first point of shadow which 
the curve of shade on the torus casts on the scotia. 

To find points of this shadow, intersect the torus and 
scotia vy a plane of rays ¢/’ perpendicular to the verti-. 
cal plane; find the point (e, e’) in which this plane inter- 
sects the curve of shade, and construct the curve in 
which it intersects the scotia. Through e draw the hori- 
zontal projection of a ray of light—the point f, in which 
it intersects the curve of the scotia, is the horizontal, 
and f’ is the vertical projection of a point of shadow. 

The shadow cast on the scotia by the curve of shade 
on the torus, terminates in the upper circle of the lower 
fillet. A part of this shadow, beginning at g, is seen m 
horizontal projection. 

The element of shade on the lower fillet is (g, 2’); 
and this element casts a shadow, beginning at its foot, 
on the lower torus. Then, that part of the upper cir- 
cle of the fillet, ntercepted between the element of shade 
and the point in which it is met by the shadow on the 
scotia, casts a shadow on the lower torus. ‘Then the 
shadow on the lower torus is. cast by the curve of 
shade on the upper torus; and this shadow continues 
until it intersects the curve of shade determined as in 
Prob. 12. 


, 


if 


£ 


74 TREATISE ON 


PROBLEM XVII. 


To find the shades and shadows on the capital and shaft of 


the Roman Dorie column. 


48. The part of the abacus, whose vertical projec- 
tion is s 7’, casts a shadow on the cyma-reversa. 

The semi-reversa is composed of parts of cylinders 
whose elements are respectively parallel and perpendi- 
cular to the vertical plane of projection. ‘Two of the 
cylinders intersect each other in a curve, whose hori- 
zontal projection is gn”, and vertical projection gpn. 

The line (rr, 77") of the abacus casts a line of shadow 
on the cyma-reversa parallel to itself. Through (7’, r) 
draw a ray of light, and find the point in which it pierces 
the surface of the cylinder whose elements are parallel 
to the vertical plane: gq‘ is the horizontal Ime drawn 
through the point, and is the lne of shadow required. 
The light falls on a part of a cyma-reversa below this 
line. Let a tangent plane of rays be drawn to the part 
of the cyma-reversa, whose elements are parallel to the 
vertical plane. It touches the cyma-reversa in the ele- 


i - ment pp’, which is therefore an element of shade. This 
‘element of shade casts a shadow on the part of the 


cyma-reversa below it. This shadow is determined by 
finding the intersection of the tangent plane of rays with 


_» the cyma-reversa. It is the horizontal line near zz’. 


The lower element x7’ of the cyma-reversa casts a 
shadow oo’ parallel to itself, onthe part of the abacus 
below it. 

We come next to the echinus, which is the half of a 
torus. 


SHADES AND SHADOWS. 15 


First, find the curve of shade on the echinus as in 
Prob. 12. Then, through the horizontal line of the 
abacus whose vertical projection is w draw a plane of 
rays: we is its vertical trace. Since this plane is per- 
pendicular to the vertical plane of projection, the shadow 
cast by the line of the abacus is vertically projected in 
the trace we’. The ray of light through the point (w”, w) 
pierces the neck of the column at the point (ce, ¢’). 

Through the lower line of the abacus (w’n”, ww’) let 
a plane of rays be drawn. ‘This plane intersects the 
echinus in a curve which is the shadow cast by the line 
(w'n, ww’) on the echinus ; this curve of shadow meets 
the curve of shade in the points x and y; leaving a part 
of the echinus in the light. The curve of shadow is 
found by intersecting the echinus by horizontal planes 
—each plane will intersect the echinus in a horizontal 
circle and the plane of rays in a right line: the point in 
which the right line intersects the circle is a point of the 
curve of shadow 

The plane of rays passed through the line (w"n”, w w’) 
intersects the neck of the column in the curve ct, the 
cavetto in the curve u, and the fillet directly above, in 
the curve v. 

The curve of shade on the echinus casts a shadow 
on the fillet below, which begins at /, and intersects at 
v the shadow cast by the lower line (w’n”, ww’) of the 
abacus. 

The lower circle ££’ of the fillet, casts a shadow on the 
echinus, which begins at a, and intersects the shadows cast. 
by the abacus in two points, one near a, the other at w. 

From uto z the shadow is cast by the small part of the 
arc of the circle Zk’, whichis in the light near v. The 
ray through z passes through the last point of the circle 
kk which is in the light, and therefore passes through a 
point of the curve of shade onthe echinus. From z to 


: 


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76 sie" ATISE ON Fy 


the line 77’, the shadow is 5 ee by the curve of shade on 
the echinus. 

The lower circle ¢7’ of the cavetto, casts a cHadows ! 
on the neck of the column which intersects at ¢ the” 
shadow cast by the lower line (w"n’”, w w’) of the abacus, 
and at ¢ the shadow, cast by the curve of shade on the «+ 
echinus; from / to?” the shadow is cast by the curve 
of shade on the echinus. - 

We come next to the astragal ff’. Find the curve -. 
,of shade as i ‘Prob:12. | A as 
' The curve of shade on the echinus casts a shadow 
on the astragal which intersects the curve of shade 
near og’. Thecurve of shade on the astragal casts a © 
shadow on the fillet dd’e’e. 

The lower circle ee’ of the fillet, casts ashadowon | 
the shaft of the colimn, which intersects the element of 
shade at 6. 

: The shadow cast by a curve of shade on asurface of 
revolution, may be found by drawing rays throughits 
different points, and findmg where they pierce the sur-  » 
face; but it is generally better to pursue the method 
adopted in Prob. 14. 


ae 

~ : 
a + * y* 
- 


OF THE HELICOID.* 


49. The helicoid is a surface generated by arightline —, 
moving uniformly in the direction of another right line 
which it intersects, and having at the same time a uni- 
form angular motion around it. 

. Therfixed line is called the axis of the helicoid, and “" « 
the moving line the generatrix. ) a 
“it te 


* The properties of the helicoid are used in the next problem, and © | 
although their discussion belongs rather to warped surfaces, than toa 
treatise on shades and shadows, yet it was thought best to give the  ~ 
properties here, as the student may not meet with them elsewhere. 


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SHADES AND SHADOWS. ool 6 | 


Let (A, BC) (PI. 11, Fig. n) be the axis of a  helicotd, 


and (DA, D’E) its generatrix. 

By the =p na Ee of the surface, the generatrix has 
two motions; one in the direction of he) axis (A, BO), 
the other around the axis, both of which are, uniform. 

Suppose the generatrix to move around the axis until 
it is horizontally projected in AF, and that during this 
revolution, it has moved in the direction of the axis a 
distance equal to EG. 

Through F draw F PY’ perpendicular to the sround 
line, and nie PE’ equal to EG—then F’G will be the 
vertical projection of the generatrix, from its new posi- 
tion; for every point of the generatrix has moved an 
equal distance in the direction of the axis. 

By considering the circumstances of the motion, it 1s 
plain: 

Ist. That every point of the generatrix, except the 
one in which it intersects the axis, describes a curve— 
the curve so described is called a helz. 

2d. That since the generatrix preserves the same 
position with the axis, all the points of the same helix 
will be equally distant from the axis, and consequently, 
the projection of any helix on a plane perpendicular to the 
axis, is in the circumference of a circle described about 
the centre A. The helix described by the point (D, D’) 
is horizontally projected in the semicircumference DIF. 

When the generatrix has moved from one position to 
another, the motion in the direction of the axis is pro- 
portional to the part of the axis which the point of inter- 
section has passed over; and all the points of the gene- 
ratrix have moved equally in the same direction. ‘The 
motion round the axis, or the angular motion of the 


a* 


generatrix, is proportional to the angle included between - 


two planes passing through the axis and the two posi- 


% 


78 TREATISE ON 


tions of the generatrix. The angle between these 
planes is measured by the arc of any circle described 
about the centre A. It may therefore be measured on 
the arc DIFR, or on the projection of any helix of the 
surface. 

Since the motion in the direction of the axis, and the 
angular motion are both uniform, their measures will be 
proportional to each other: that is, when the point E 
of the generatrix has passed over a part of CB, the 
angular motion of the generatrix is measured by a like 
part of DIF. . 

Let the revolution of the generatrix (AF, GF’) be 
continued until the point (F, F’) shall be horizontally 
projected at D. The point G will then have moved to 
C, and the point (fF, F’) to (D, D”): GC and KD" being 
each equal to EG. The revolution of the generatrix 
may be continued at pleasure. 

The horizontal projection of the helix described by 
the point (D, D’) is the circumference DIFR. It is 
required to find its vertical projection. 

Let the semicircumference DIF be divided into any 
number of equal parts, say eight; and divide the dis- 
tance D’'K into the same number of equal parts, and 
draw the parallel lines as in the figure. 

Now, when the point (D, D’) is horizontally project- 
ed at a, it will have ascended one-eighth of the distance 
D’K, and is, therefore, vertically projected ata. Hence 
ad is a pomt in the vertical projection of the helix. 
Each of the other points of division also gives a point, 
and the vertical projection of the helix, from D' to F’, 
can be drawn; the part F’D" is easily found, since it 
has the same position with the horizontal plane KF’, 
as the part D‘a’E’ has with the plane D’P. The part 


SHADES AND SHADOWS. 79 


D’X has also the same position with the horizontal 
plane D"E, as the part D’a’ fF’ has with the plane:D’P. 

The curve DQN, in which the horizontal plane of 
projection intersects the surface, is determined by find- 
ing the points in which the elements of the surface 
pierce the plane. This curve is the spiral of Archi- 
medes—it extends indefinitely, unless we suppose the 
revolutions of the generatrix to be limited. 

If through the poimt (D, D’), the line (DA, D'T) be 
drawn, making an angle with the axis of the surface equal 
to the angle D'EB, we may regard this line as the genera- 
trix of a helicoid, similar:in every respect to the one 
already described, excepting that it has a different posi- 
tion with the horizontal plane. We shall, for the sake 
of distinction, call the first the upper hehcoid, and the 
second the lower helicoid. ‘The helix described by the 
point (D, D’) is common to the two surfaces. Regard- 
ing the generatrices as indefinite, the surfaces are inde- 
finite helicoids. 

The helicoid is a warped surface, since the consecu 
tive positions of the generatrix are not in the same 
plane (Des. Geom. 72). 

tt ts now required to draw a tangent plane to the lower heli- 
coid, at a point of the common helix. 

50. Let (D, D’) be the point at which the tangent 
plane is to be drawn; and let us find its trace on the 
horizontal plane SF’, drawn at a distance above the 
point of contact (D, D’), equal to the ascent of the 
generatrix in half a revolution. 

The tangent plane must contain the element (DA, 
D'T), (Des. Geom. 89); and therefore, (S, S’) where it 
pierces the plane S’'F” is one point of the trace. But the 
tangent plane must also contain the tangent line to the 


80 TREATISE ON 


helix at the pomt (D, D’) (Des. Geom. 88); and these 
two lines determine its position. 

The helix (DaF, D’a'F’) is on the surface of the right 
cylinder which projects it on the horizontal plane. Con- 
ceive a tangent plane to be drawn to this cylinder along 
the element which pierces the horizontal plane at D. 
The tangent line to the helix at (D, D’) will be con- 
tained in this plane. 

Let the surface of the right cylinder be developed on 
the tangent plane. ‘The semicircumference DIF will 
develop into a right line (Des. Geom. 131); and since 
the arcs Da, Dd, &c, are. proportional to the dis- 
tances of the corresponding points of the helix above 
the horizontal plane, it follows that the helix will de- 
velop into a right line. 

Then, lay off DV equal to the semicircumference 
DIF, and draw the perpendicular VF”, and make it equal 
to PF’: we have then DF” for the development of the 
helix. But the tangent line at (D, D’) and the helix 
make the same angle with the horizontal plane; there- 
fore, the helix developed coincides with the tangent line 
at (D, D'). Hence, the tangent line to the helix at the 
point (D, D'), pierces the horizontal plane SF’ at the 
point (V, K), a distance from the plane DA equal to the 
semicircumference DIi'. Therefore, SV is the hori- 
zontal projection of the trace of the tangent plane 
on the plane S'I’. 

51. It is evident, from the similarity of triangles, that 
if the horizontal plane on which the trace is found, be 
taken above or below SEF’, the distance D’K will be to the 
semicircumference DIF, as the distance from D’ to the 
plane, is to the distance from the vertical plane DAF to 
where the tangent pierces the horizontal plane. There- 
fore, if the horizontal plane S'F’ were passed through D’, 


SHADES AND SHADOWS. 8] 


making the distance from D’ equal to the ascent of the 
generatrix during an entire revolution, the distance DV 
would be equal to the entire circumference DIFR. 

52. By considering the generation of the helicoid, it 
is evident that each helix makes a different angle with 
the horizontal plane; the limits of these angles being 
90° and 0. 

For, the helix described by the poimt in which the 
generatrix intersects the axis, is the axis itself, and is 
therefore perpendicular to the horizontal plane. The 
helix described by the point at an infinite distance from 
the axis may be considered horizontal, since the dis- 
tance which the point moves in the direction of the 
axis is inconsiderable in comparison with its angular 
motion. 

Since every tangent plane to the helicoid passes 
through an element (Des. Geom. 89), the limits of the 
angles which tangent planes make with the horizontal 
plane are 90°, and the angle made by the generatrix. 


PROBLEM XIX. 


Having given the projections of a screw and the direction 
of the light, tt 1s required to find the shades and shadows on 
the different parts of the screw, and the shadow cast on a hori- 
zontal plane. 


53. Let there be a vertical cylinder whose axis is 
(A, A'B), (PI. 11) and the radius of whose base is AC. 
Let the base of an isosceles triangle, whose vertical 
projection is C’D‘E, be placed to coincide with the ele- 
ment of the cylinder which pierces the horizontal plane 
at C; the vertex D’ of the triangle being in the hori- 
zontal plane at D. The plane of the triangle will pass 
F 


“eS .. a 


82 “ PREATISE ON 
through the axis of the cylinder, and therefore the two 
equal sides will both intersect it. 

Let the triangle be now revolved about the cylinder, 
having, at the same time, a uniform motion in the direc- 
tion of the elements—the plane of the triangle con- 
tinuing to pass through the axis. 

When the triangle has been revolved half round the 
cylinder, suppose it to have ascended a distance equal 
to half its base. The vertex (D, D*) will then have the - 
position (G,G’); G’G’ being equal to half the base. 
After the remaining half of the revolution is completed, 
the vertex of the triangle will have ascended a distance 
equal to its base, and will be vertically projected at F; 
FEH will then be the vertical projection of the gene- 
rating triangle. 

The cylinder about which the triangle has been re- 
volved, is called the cylinder of the screw. 

The solid generated by the triangle, in an entire revo- 
lution, is called the thread of the screw. The surface 
generated by the upper equal side of the triangle is 
called the upper surface of the thread; andthe surface 
generated by the other equal side is called the lower 
surface of the thread. ‘These surfaces are portions of 
helicoids (49). The helicoids may be considered inde- 
finite, by supposing the equal sides of the triangle to be 
indefinitely produced. 

The helix described by the vertex (D, D’), is called 
the outer helix of the screw; and the one described by 
either extremity of the base, is called the inner helix 
The curves described by the pAb HOC: points are 
called helices. 

The vertical projections of the outer and inner helices 
are constructed as in Art 49. The parts which lie in 
front of the plane DAG are made full, those which lie 


- * a 


oe ° 
SHADES AND SHADOWS. 83 


behind it are dotted. The horizontal projection of the 
outer helix is the circumference described with the 
radius AD, and of the inner helix, the circumference 
described with the radius AC. 

The outer helix rises from the horizontal plane at D, 
and passes behind the plane DAG at (G,G’). The 
inner helix rises above the horizontal plane at I, and 
passes in front of the plane DAG at (C, E). 

If the triangle EFH be again revolved around the 
cylinder of the screw, it will generate a second thread; 
and every new revolution will give an additional thread. 

The base of the triangle, or the distance which it has 
ascended in an entire revolution, is called the distance 
between the threads. 

The curve into which the surface of the screw inter- 
sects the horizontal plane is found as in Art. 49, except- 
ing that the elements are differently inclined to the hori- 
zontal plane, which, however does not vary the prin- 
ciples of the construction.—It is the curve Dal. 

The screw is usually worked into a block of wood 
called a fillet. We have taken the fillet octagonal; the 
octagon 4 1 2 3 5 is its horizontal projection, and the 
- rectangle above the threads its vertical projection. 

It is now required to find the curve of shade on the 
lower surfaces of the threads. 

Let a plane be drawn tangent to the lower surface of 
_ the thread, at the point (D, F) (50), and let its trace be 
constructed on the horizontal plane L’H, taken at a 
distance above the point of contact equal to the half 
distance of the threads. 

The point (L, L’), in which the element passing 
through the point of contact pierces the horizontal 
plane L’H, is one point of the trace of the tangent 
plane. Drawing through ‘ the perpendicular DO, and 

4 ee . 


84 TREATISE ON 


making it equal to the semicircumference DaG, we have 
the distance from the plane DAG, at which the line 
tangent to the helix at (D, F) pierces the horizontal 
plane L'H (50). This distance should have been laid 
off in front of the plane DAG, but this could not be 
done, for want of room on the paper. If however, we 
suppose a line drawn through L and the point O, taken 
at a distance equalto DO in front of the plane DAG, it 
will make an angle with LD equal to the angle DLO. 
Hence, if Lk’ be drawn, making the angle DL£’ equal 
to the angle DLO, it will be the horizontal projection of 
the trace of the tangent plane. ‘The tangent plane cuts 
the axis of the screw at (A, I’). 

Now, if this plane be a plane of rays, the point of 
contact (D,I*) will be a point of the curve of shade. 

To ascertain if it be a plane of rays, draw through 
(A, I’) a ray of light; its projections are AK and IK’, 

If the tangent plane be a plane of rays, the ray of 
light having one point in common, will coincide with it ; 
and if it comcide with it, it will pierce the horizontal 
plane L'H in the trace &L#’. But it pierces this horizon- 
tal plane at (K, K’) out of the trace—therefore, the 
tangent plane is not a plane of rays. 

Let the ray (AK, IK’) be revolved about the axis of 
the screw; it will generate the surface of a right cone 
whose vertex is (A,I’), and the point (K, K’) will describe, 
in the horizontal plane K’L’‘H, the arc of a circle of 
which £K£’ is the horizontal projection. The tangent 
plane to the screw will intersect the surface of this cone 
in two elements whose horizontal projections are 
Ak and AF’. 

These lines may be regarded as the ray (AK, IK’) 
after it has been revolved about the axis of the screw 
to coincide with the tangent plane. 


_ SHADES AND SHADOWS. | 


Let the tangent plane be now revolved around the 
axis of the screw, in such a manner as to continue tan- 
gent along the outer helix, until the revolved ray Ak 
shall be horizontally projected in AK. In this revolution 
of the tangent plane, every point of the line Ak will 
ascend equally; therefore, after the counter revolution 
the vertical projection of the line will be parallel to IK’; 
hence, the line itself will be a ray of light. 

In the revolution of the tangent plane, every point 
has an equal angular motion, and this motion is mea- 
sured by the are oDé; the point of contact (D, F) having 
the same angular motion, if we lay off Déa equal to 
oDé, or what is the same, ba equal to oD, we find a, the 
horizontal projection of the point of contact when the 
tangent plane is a plane of rays. Projecting a into the 
vertical projection of the outer helix, we have one point 
(a, a’) of the curve of shade. 

If we revolve the tangent plane from D towards 0, as 
we are at liberty to do, causing it to descend along the 
outer helix but continuing tangent to the surface, until 
the line AZ’ shall be horizontally projected in AK, the 
plane will, for the reasons given above, become a plane of 
rays, and the point of contact,a point of the curve of 
shade. The angular motion, in this revolution, will be 
measured by dc—hence, laying off De’ equal to dc, we 
have c’, the horizontal projection of the point of contact. 
Projecting the point c’ into the vertical projection of the 
outer helix at c’, determines (c, c’),a point in a second 
curve of shade. Hence we see that two tangent planes 
of rays can be drawn touching the same thread of the 
screw in points of the outer helix. 

Let a tangent plane be drawn to the under surface of 
the thread of the screw, at the point (C, E) of the inner 
helix; and let its trace be constructed on the horizontal 


86 TREATISE ON : 


plane K’L’‘H, at a distance above the point of contact 
equal to the ascent of the generating triangle in an entire 
revolution. The point (L, L’), in which the element 
through the point of contact pierces the horizontal plane 
L’H, is one point of the trace, and making CP equal to 
the circumference Cef If (51), and considering P in front 
of the plane DAG, we have a secondpoint of the trace, it 
being the point in which the tangent line to the helix at 
(C,E) pierces the plane L'H. Drawing LP, and then 
the line Ld’, making the angle DL’ equal to the angle 
DLP, we determine d’Ld the trace, on the plane L’'H, of 
the tangent plane to the inner helix at the point (C, E). 

This tangent plane cuts the axis of the screw at the 
point (A, I’), and since the ray through this point does 
not pierce the horizontal plane K’H in the trace of the 
tangent plane, it follows, that the tangent plane is not a 
plane of rays. Let the ray (AK, IK’) be revolved about 
the axis of the screw as before—it will generate the sur- 
face of a right cone, which the tangent plane will inter- 
sect in two elements whose horizontal projections are 
Ad and Ad’. 

The tangent plane is now revolved to become a plane 
of rays, as in the last case. ‘The angular motion, when 
it ascends, is eCg, and when it descends, is fg. There- 
fore, making ge’ equal to Ce, and Cf’ equal to gf, we have 
é and f’ the horizontal projections of the two points of 
tangency. ‘The point ¢ is vertically projected on the 
inner helix at ¢’, and the point f’ at f’—hence, two 
points in each of the curves of shade are found. It is 
now required to find intermediate points. 

Any element of the lower surface of the thread, whose 
horizontal projection passes between the points a and é, 
will contain an intermediate point of the curve of shade. 


XM 
ae 
+ 


SHADES AND SHADOWS. 87 


Assume Ah for the horizontal projection of such an 
element. 

Projecting the point / into the outer helix at h’, and 
the point ¢ into the inner helix at 2’, we find Rz‘h’ the 
vertical projection of the element. 

Through (A, h'), the upper extremity of the element, 
draw the horizontal plane M’h’, and through the point 
(A, R) draw a ray of light (AM,RM’); this ray pierces 
the horizontal plane M’h’ at the point (M, M’). 

If through the element (AA, Rh’) we suppose a plane 
of rays to be passed, its trace on the plane M7’ will be 
the line of which Mh is the horizontal projection; and 
since the surface of the screw is a warped surface, this 
plane will be tangent to it at some point of the element 
(Des. Geom. 229): the point of contact is a point of 
the curve of shade. 

Let us suppose for a moment this point of contact to 
be found, and that it is (u, uv’). At a distance above the 
point of contact equal to half the distance of the 
threads, let the horizontal plane N’H’ be drawn. The 
trace of the plane of rays on this plane is parallel to its 
trace on the plane M7’; therefore, NA’ drawn parallel 
to Mh is the trace on the plane N’H’. 

If then, through the point u, a perpendicular be drawn 
to Au, and produced till it meets Nh” in Q, the distance 
uQ will be the semicircumference of the circle whose 
radius is Au (50). But h’w is the horizontal projection of 
a part of the element intercepted between two horizontal 
planes, at a distance from each other equal to the half 
distance of the threads ; and since all the elements make 
the same angle with the horizontal plane, this projection 
is equal to the projection of D’E, that is, to DC or hié. 
Hence, if through 7, 2S be drawn perpendicular to Azh, 
we shall have two similar triangles, h’vwQ and hiS, having 


™. 


88 TREATISE ON 


in each an equal homologous side h” u and he; hence the 
sides uQ and 7S are equal. But the line vwQ has been 
proved equal to the semicircumference of the circle 
passing through the point of contact; therefore, 2S is 
also equal to this semicircumference. 

After having drawn the trace MA, of the plane of rays, 
we have only to draw 7S perpendicular to AzA and find the 
tadius of the circle of which 2S is the semicircumference. 
This is easiest done by laying off 2S from D, on DO, and 
drawing through the extremity of the line a parallel to 
the line joining A and O—the distance cut off from D, on 
DA, will be the radius required; for the semicircumfer- 
ences are to each other as their radii. ‘Then, with this 
radius, and A asa centre, describe an arc; the point u, 
where it cuts Ah, is the horizontal projection of the 
point of contact, and this point bemg projected into the 
vertical projection of the element atu’, the vertical pro- 
jection of the point of contact is also determined. An 
intermediate point in the curve of shade on the other 
side of the plane DAG, may be found by a similar con- 
struction. 

Since the light has the same position with the lower 
surfaces of all the threads, the curves of shade upon 
them will be directly over each other: therefore, if a 
vertical line be drawn through (a, a’), the points in which 
it cuts the outer helices will be points of the curves of 
shade, and the vertical line through (e¢, e’), will deter- 
mine corresponding points of shade on the inner helices. 

_ It isnow required to find the shadow which any thread 
will cast on the surface of the thread directly below it. 
The curve of shade, and a part of the outer helix, 
will cast shadows on the thread below them. The sha- 
dow will begin at the point where the curve of shade 
meets the inner helix. 


SHADES AND SHADOWS. 89 


To find the shadow cast by the curve of shade, we 
first find the shadow which the curve of shade would 
cast on the horizontal plane if the screw were removed. 
This is done by drawing rays of light through its seve- 
ral points, and finding where they pierce the horizontal 
plane. We then take any element of the upper surface, 
near to (¢’, ¢’), on which we suppose a shadow will fall, 
and find the shadow which it would cast on the horizon- 
tal plane. Through the point p, where these shadows 
intersect, 1f we suppose a ray of light to be drawn, it 
will intersect both the element and the curve of shade. 
The point where it meets the element is the point of 
shadow on the element, and the point where it meets the 
curve of shade is the point casting the shadow. The 
ray through p gives (p’, p’) for the point of shadow. 

To find the shadow cast by the helix. Take any ele- 
ment on which we suppose the shadow will fall, and 
through it pass a plane ofrays. Find the point in which 
this plane cuts the outer helix, directly above, and 
through this point draw a ray of light: where it inter- 
sects the element is a point of the required shadow. 

Let Ag be the horizontal projection of an element on 
which it is supposed the shadow will fall, andg v its 
vertical projection. ‘Through (q, q’), the foot. of the ele- 
ment, let the horizontal plane qT’ be drawn. ‘Through 
(A, v), the point in which the element intersects the axis, 
let a ray of light be drawn; this ray pierces the horizon- 
tal plane through (q, q/) at (T, 'T")—therefore gT is the 
trace, on the plane ql", of the plane of rays through 
the element. It is now required to find the point m 
which this plane of rays cuts the outer helix (DagG, FX.) 

Through the axis of the screw let a plane be 
passed perpendicular to the plane of rays—its hori- 
zontal trace is Ar perpendicular to gT. Let the plane 


OG TREATISE ON 


of rays and the plane whose horizontal trace is Ar be 
revolved about the axis of the screw until Ar becomes 
parallel to the ground line; and suppose.the plane of 
rays to have at the same time an ascending motion in 
the direction of the axis, such that (Ag, vq’) shall con- 
tinue an element of the screw. When Ar is parallel to 
the ground line, the plane of rays will be perpendicular 
to the vertical plane of projection, and the angular 
motion will be measured by the arc rG. 

From qg laying off gq” equal to rG: Ag” will be the 
horizontal projection of the element after the revolution, 
and g"s' will be its vertical projection. But, since the 
element is a line of the plane of rays, which, in its re- 
volved position is perpendicular to the vertical plane, 
the projection of the element is the vertical trace of the 
plane (Des. Geom. 24). 

Therefore s’, the point in which the line q's’ inter- 
sects I°X, is the vertical projection of the point in its 
revolved position, in which the plane of rays cuts the 
outer helix. The horizontal projection of this point is 
ats. In the counter revolution, the point (s, s’) has the 
same angular motion as the other points of the element 
—therefore, laying off ss” equal to rG, and projecting s” 
into the outer helix at s, determines (s”, x”) the point 
in which the plane of rays cuts the outer helix. 

Through this point draw a ray of light—the point 
(n, n'), where it intersects the element, is a point of the 
curve of shadow. By similar constructions we may find 
other points of the curve. Having found several points 
of the curve of shadow, let its two projections e'p'nt, 

e"p'n't’ be drawn. 

The shadows on the threads HOE above, are 
similar to the one already found, and are all horizon- 


tally projected in the curve e’p'nt. 


tit 


SHADES AND SHADOWS. 9j 


Drawing through ?¢ the vertical projection of a ray 
of light, the point /, where it meets the outer helix above, 
is the vertical projection of the point which casts the 
shadow at (¢,¢’); and a’‘l is the vertical projection of 
the part of the outer helix which casts a shadow on the 
thread. 

There are similar shadows on the other side of the 
plane of rays KAT’, which are constructed in the same 
manner as those already found. 

If through any element of the upper surface of a 
thread, a plane of rays be drawn, and its trace con- 
structed on the horizontal plane through the foot of 
the element, the horizontal projections of these traces 
will all pass through the point T. 

For, suppose we take the element whose vertical pro- 
jection is g’’s’. Through g” draw the horizontal plane 
qT". The ascent of the foot of this element,in revolving 
from the position (Ag, vq’),1s equal to T’T”, and the point 
v has ascended the same distance. Therefore the ver- 
tical projection of the ray drawn through the point in 
which the element intersects the axis, will intersect the 
vertical line T’T” at T”, which is in the horizontal plane 
gT”; and as the same may be shown for all other 
elements, it follows, that the horizontal projections of 
all the traces will pass through the point T. 

To find the shadow cast on the screw by the fillet: 

The lower lines of the fillet which are towards the 
source of light, and whose horizontal projections are 
12, 23, and 14, will cast shadows on the screw. 

Let it be required to find the shadow cast by the line 
whose horizontal projection is 12. 

Through this line suppose a plane of rays to be passed 
—the curve in which it intersects the surface of the 
screw is the curve of shadow required. The plane of 


mw 


92 TREATISE ON 


rays through the axis of the screw, cuts the lower line 
/12 of the fillet at the point (y¥'). ‘Through y/ draw the 
ertical projection of a ray—the point y”, where it inter- 
sects the axis of the screw, is the point in which the 
plane through 12 cuts the axis. 

Every plane passing through the axis of the screw will 
intersect the surface in an element, and the plane of rays 
in a right line passing ‘through the point (A, y”); the 
point in which these lines intersect is a point of the 
shadow. 

The plane whose trace is KA intersects the thread on 
which the shadow falls in the element (dg, ’g’), and the 
plane of rays in the line whose vertical projection 7/7"; 
hence their point of intersection is a point of the curve 
of shadow. The plane LA gives the point of shadow 
on the element parallel to the vertical plane. 

The plane of rays through 23, cuts the axis of the 
screw at X”. Having found a sufficient number of points, 
let the vertical projection of the curve be drawn. The 
shadow passes off the thread at m. The horizontal pro- 
jection is easily found, but is not made, lest it should 
render the figure on the horizontal plane too complicated. 

To find the shadow cast by the screw on the hori- 
zontal plane : 

This shadow lies on both sides of the plane KA; but 
as the construction for the two parts isthe same, it will 

only be necessary to make it for one of them. 

The shadow on the horizontal plane begins at the 
point z, where the curve of shade on the thread inter- 
sects the horizontal plane.. The curve of shade between 
this point and the outer helix, casts a shadow on the hori 

zontal plane; this shadow is found by drawing rays 
through the points of shade, and finding where they 
pierce the horizontal plane. “ 


& SHADES AND SHADOWS. | 93 


“The part of the outer helix, between the points a” and 
¢,next casts a shadow. ‘This shadow is found by draw, 
ing rays through the points of the helix, and finding’ 
where they pierce the horizontal plane. ‘The point (¢, ¢’) 
casts its shadow at ¢”, and zt” is the shadow cast by the 
curve of shade and outer helix. 

The ray through (¢, ¢’) intersects the outer helix of 
the next higher thread at. The point whose vertical 
projection is / casts ashadow on the screw at (¢,?’), and 
on the horizontal plane at ¢”. The part of the helix 
from / to the point where it intersects the shadow on the 
thread, will cast a shadow on the horizontal plane, which _, 
is found by drawing rays through the points of the helix. 

We then pass to the helix of the next higher thread, 
and so on until we arrive at the thread on which the 
shadow of the fillet falls. 

Having before found m, the vertical projection of the 
point in which the shadow of the fillet intersects the 
outer helix, we find the shadow of this point on the hori- 
zontal plane which is at m’. At this point the shadow 
of the fillet on the horizontal plane begins. This 
shadow is found too easily to require particular expla- 


nation. 
54. The brilliant point of the surface of the helicoid, 


is found by bisecting the angle between a ray of light 
and a line drawn to the eye, and then drawing a plane 
perpendicular to the bisecting line and tangent to the 
surface. The details of the construction are left as an 
exercise for the student. ise 
55. In all the constructions which have been made, 
the rays of light have been supposed parallel. It may 
be well to consider how the constructions would have 
been made had the rays been divergent or convergent. 
If we suppose the rays of light to emanate from a 


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opaque body will be the line of shade. 


ee 4 of light, wall be the indefinite shadow. When the opaque 
.. body is the largest, the luminous body and the vertex. 


“. gays of light may be considered as convarayas to the 


vw t ne ° 
des » TREATISE ON “ 


luminous point at d finite distance; and to fall upon an 


opaque body, the rays will be,divergent. 
_ Let us suppose'a cone to be drawn tangent to the — 


“opaque body, of which the Hane point shall be the - 


vertex. It is plain, 
1°. That the line of contact of fae cone with the 


eat alk r 
‘ oe a 
¢ we 


2°. That the part of Space included within the sur- 
facevof this cone, and lying on that side o “the opaque 
body opposite to the source of light, will be the indefi-_ 
nite shadow of the opaque body. a Tae 

3°. That the shadow of the opaque body on any sur-) .* 


“ig 


~ face, will be the intersection of that surface with the = 


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tangent cone of rays. 

56. If we suppose the light to emanate froma  lumin- 
ous body, we may Agee a cone drawn tangent to the 
luminous body, and to any opaque body whose shade 
~~ and shadow are to be determined. The curve of con- 


> 


| 


_tact on the opaque body willbe the line of shade, and the 


part. of space within the surface of the tangent cone 
ad on the side of the opaque body opposite the source 


of the tangent cone will be on the same side of the 

opaque - body, and the rays of light will be divergent; but 
when it is the smallest, the opaque body will be between | 

‘the luminous body’and the vertex of the cone, and the’. 


. * 
os 2 
fh 
: 4 4 


“?, 
shila of the tangent,cone. 

* In all the cases in which the rays F light are not... 
T pBeattel, the. problems in Shades and Shadows will be 
‘ solved by finding the contact of a cone with an opaque 
body, and the intersection of this cone with any surface - 
on parcnich the shadow falls. er" 


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CHAPTER L. arate Wish 


7. Hap we no knowledge’ of abe other than what 
is derived through the medium of sight, we should sup- 
pose them to differ from each other in two respects only 
—form and colour. Objects having different forms, or 
different colours, produce different effects upon the 


eye; but objects of the same form and colour cannot be 


distinguished | from each other without ‘the yaid of the 
other senses. « 


58. When we view an uate’ all the matt fa it which | 
are seen are supposed either to emit or reflect rays. of 


light which fall upon the eye; and it is through the 
medium of these rays that we derive the idea both of 


its form and colour. . ® 


59. Perspective is the art of representing objects on 


a surface, in such a manner that the representations 


shall present to the eye, situated at a particular point, 


the same appearance as is presented: by the objects 
themselves. The representation of an object so made - 


is called its perspective. a 
_*60. Let us now suppose that we are viewing an i Etiect 


in space, and that a transparent pic) is tic between. 


us and the object. * 
Every point of the object which is seen, is supposed: 


_ to emit a ray of light that falls upon) the eye, and each 


. »_ ad “. 
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96 TREATISE ON 


ray pierces the transparent plane ina point. If to each 
point so determined, a proper colouring be. given, the 
representation or picture, on the transparent plane, will 
present to the eye the same appearance as the object 
itself. Such representation is, therefore, the perspective 
of the object. j 

61. The rays of ight coming from the different 
points of the object to the eye, are called vesual rays ; 
“and the plane on which the representation is made, is 
called the Perspective Plane. 

62. The art of Perspective is therefore divided into 
two parts. ' 

1°. To find the points in which the visual rays pierce 
the perspective plane, which determines the general 
outline of the perspective. 

2°, So to shade and colour this outline, that it shall 
appear in every respect like the object itself. 

The first part is called Linear Perspective. It em- 
braces that portion of perspective that is strictly mathe- 
matical, and which will form the selltthy of the following 
treatise. 

The second part is called Merial Perspective. This 
branch of the art belongs to the draftsman and the 


painter, and is to be learned by a careful study of the © 


objects of nature, under the guidance of an improved 
and cultivated taste. 

63. As it is the end of perspective to represent objects 
as they appear in nature, such a position ought to be 
given to the perspective plane as will enable us to con- 
ceive, most easily, of the positions of objects from 
viewing their perspectives. 

This we can do with the least difficulty, when the per- 
spective plane is taken parallel to the principal lines of 
the eis . 


ob 


LINEAR PERSPECTIVE. 97 


Of the objects in nature, the larger portion of lines 
*4 are vertical; therefore, in Tost perspective drawings 
the perspective plane has a vertical position. 
64. Before an object can be put in perspective three 
things must be given, or known. 
1°. The place of the eye. 
2°, The position of the perspective plane. 
. The position of the object to be put in perspec- 
aa | 
65. By considering what has already been said, we 
may deduce the following principles : 
1°. If through the eye and any point in space, a visual 
ray be drawn, the place at which it pierces the perspec- 
tive plane is the perspective of the point. 
2°, If through the eye and all the poits of a right 
line, a system of visual rays be drawn, they will form a 
plane passing through the eye and the right line; this 
plane is called a ‘Sel plane, and its trace on the per- 
spective plane is the perspective of the right line. 
3°. If through the eye and all the points of a curve, 
a system of visual rays be drawn, they will, in general, 
form the surface of a cone, the vertex being at the 
eye; this cone is called a visual cone, and its intersec- 
tion with the perspective plane is the perspective of the 
curve. Oot 3 
66. In determining the perspective of an object, it is 
unnecessary, and indeed impracticable, to draw visual 
rays through all its points that are seen, and to con- 
struct the intersections of these rays with the perspective 
»plane. We therefore select the prominent points and 
lines only, such as the vertex and edges of a pyramid, 
the vertex of a cone, the edges of a prism, &c.: and 
having put these lines in perspective, we have, in fact, 
determined the pepepectrmer the body. 


% 


98 TREATISE ON 


67. If through the eye a system of visual rays be 
drawn tangent to the object to be put in perspective, they 
will, in general, form the surface of a visual cone tan- 
gent to the object; the line of contact is called the ap- 
parent contour of the object; and the intersection of the 
surface of this cone with the perspective plane, is the 
boundary of the perspective of the object. 

We shall now apply these principles in finding the 
perspectives of objects. 


PROBLEM I. 


Having given a cube and tts shadow on the horizontal 
plane, it ts required to find the perspective of the cube and the 
perspective of ats shadow. 


68. Let DEGF (PI. 12) be the horizontal projection, 
and D'E'G’E” the vertical projection of the cube. And 
let DehgG be the shadow cast on the horizontal plane. 

Let (A, A’) be the place of the eye, and BC, BC’ the 
traces of the perspective plane. 

The perspective of the cube and its shadow, after 
they shall have been found, will be projected on the 
planes of projection in the traces of the perspective 
plane (Des. Geom. 24); but in order to exhibit their per- 
spective truly to the eye, it must be presented as it 
appears on the perspective plane. 

For this purpose we remove the perspective plane 
parallel to itself, any convenient distance, as BB’, and 
then revolve it about its vertical trace B’E’ until it coin- 
cides with the vertical plane of projection. The place 
of the eye is supposed to be moved with the perspective 
plane, and to have the same relative position with it 
after it has been revolved. 


LINEAR PERSPECTIVE. 99 


Through the angle (D,D’) of the cube, draw the visual 
ray (AD, A’D’); this ray pierces the perspective plane 
at the point (d, a’), and after the plane has been moyed 
and revolved, the perspective of the point is at d”: on 

Through (E, D’) draw the visualray (AE, A’D’) ; nis 
ray pierces the perspective plane at (a, a’), and deter- 
mines a” the perspective of (E, D’). Hence d”a” is the 
» perspective of DE. 

The*visual ray through the point (F, F’) pierces the 
perspective plane at the point (/", /’), and determines 
e, the perspective of the point (EE). The visual ray 
through the point CE, E’) pierces the perspective plane at 
(a, is and determines £’, the perspective of the point 
(E, E’). Hence ak’ is the’ perspective of the edge 
(E, D’E’) of the cube. 

The visual ray through (D, E’) pierces the perspective 
plane at (d, £), and determines p, the perspective of the 
_ point (D, E’). Hence pk’ is the perspective of the edge 
(DE, E’) of the cube, d"“p of the edge (D, D’'E’), and 
the square a’k'pd” of the front face of the cube, which 
is parallel to the perspective plane. : 
Having determined, by similar constructions, the per- 
_ spectives of the other angles of the cube, we see that 
the trapezoid d"pq'f is the perspective of the face which 
is horizontally projected in the line DI’; the trapezoid 
pkn'd the perspective of the upper face of the cube; 
the trapezoid d"a"n'f the perspective of the base of the 
~ cube; the square fn'n’d the perspective of the back face of 
the cube which is parallel to the perspective plane; and 
the trapezoid a’k'n'n' the perspective of the face of the 
cube which is horizontally projected in the line EG. 

There are but three faces of the cube which are seen, 
viz. the upper face, the front face parallel to the per- 


spective plane, and the face which is horizontally pro- 
G2 . 


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100 , TREATISE ON eet 


jected in the line FD. ©The perspectives of the ines — 
ee ee faces are made full in the perspective 

ne. e perspectives of the other edges are dotted, 
in order to show how the cube would appear if it were 
a transparent body. 

The perspective of the shadow on the horizontal - 
plane is found by finding the perspectives of the elas 

e,h, and g, and tins the lines d"e, eh, hg and gn. 2 

The jie hg, which is the perspective of hg, inter- 
sects at c the line fg,which is the perspective of the 
edge (F, FG’) of the cube. The part eg of the line hg 
not seen. 

[f through the vertical edge (1°, F’G’) of the cube, a 
visual plane be passed, it will cut the line hg in the point ~ 
whose perspective isc. ‘That part of the line hg lymg. % i 
between the visual plane and the vertical plane of pro- . es 
jection will not be seen, because the cube intervenes. 

The visual ray passing through the poimt whose per- 
spective is ce, will intersect the line hg, and the ose i 
(F, I"G") of the cube; and generally the visual ray pass- » ©. 
ang through the point inwhich the perspectives of two lines. 
ee will intersect both the lines in. space. 

69. This method of perspective is often used advan- ~ 
_ tageously in finding the perspectives of bodies bounded 
by curved foriees. If through the eye a plane be. 
passed tangent to the object to be put in perspective, the 
point of contact will be a point of the apparent con- ©. ~ 
tour of the object’ (67), and the visual ray drawn * 
through the point of tangency will determine a point in ~ 
the boundary of the perspective. For example, if it 


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LINEAR PERSPECTIVES 101 ry 


then draw visual rays through the points of _contact ; pee 
the points in which these visual rays, pierce the "per- a 


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70. The point from which the eye is supposed to view . ear 

* an object put in perspective, 1s called the point of si sight; 

and the projection of this ; point on the perspective plane. ca 
4 is called the centre of the picture. © —_s ere 
d P 71. If through the point of sight a right line be drawn aM 
> parallel to any right line in space, the point in which it en 
* .« pierces the:perspective plane is — the Ciheolh & point 
, of that line. # 

Hence, all parallel right lines née the same vanishing < 
point : for a right line parallel to one of them will be ad 
parallel to all the others. . . The % 

Hence also, all lines perpendicular to the perspective, 
plane have their vanishing pone at the. centre*of-the#,’.. .,7¢ 
picture. wl is cabs a NEED ic 

Regarding a right line as indefinite i in length, i its\van Ne ga 
ishing point is a point of its perspectives; For, the par- S ae 
allel through the point of sight is contained in the visual ee 
plane passing through the given line; therefore, the point . 
at which it pierces the perspective plane is in the trace of. 
of the visual plane. But the trate of the visa pane a es 


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102 -«s “TREATISE ON 


is the perspective of the given line (65); therefore, the 
vanishing point of a line is a point of its perspective ; 
and is the perspective of the point at an infinite distance 
from the perspective plane. 

_ When <a line is indefinite in length, its perspective 
is called the indefinite perspective of the line. 

It follows from what has just been shown, that the 
centre of the picture 1s common to the perspectives of all lines 
which are perpendicular to the perspective plane. 

72. The point in which a line pierces the perspective 
plane, is also a point of its perspective. When there- 
fore we have found this point, and the vanishing point, 
the perspective of the line can be drawn. 

If a system of lines be parallel to the perspective 
plane, the line drawn through the point of sight parallel 
to them, will also be parallel to»the perspective plane; 
and hence, the vanishing point of the system will be at 
an infinite distance from the centre of the picture. 

Now, since the perspectives of all the lines must pass 
through their vanishing poit, and since their vanishing 
point is at an infinite distance from the centre of the 
picture, it follows, that the perspectives of lines so 


situated are parallel to each other. We can prove in 


another way that these perspectives are parallel. 

For, the visual plane passing through either of the 
parallel lines, will intersect the perspective plane in a 
line parallel to the system of lines; hence, the traces of 
the several visual planes will be parallel to each other. 


But these traces are the perspectives of the given lines; 


therefore, when a system of parallel lines is parallel to 
the perspective plane, their perspectives will be parallel 
to each other. . 

It follows, from what has been said, that. when a line 
1S parallel to the perspective plane, the perspective of 
the line will be parallel to the line itself. 


$e 


ee on eee 


LINEAR PERSPECPIVE. 103 


73. If we have a system of parallel rightJines, which 
are not parallel to the perspective plane, their perspec- 
tives will meet in a common point. For, the system of 
parallel lines will have a common vanishing point (71); 
and the indefinite perspective of each hne will pass 
through this point. 

The same fact may also be proved otherwise. For, 
suppose visual planes to be drawn through the lines. 
These visual planes will all intersect each other in a line 
passing through the eye and parallel to the system of 
lines; the point in which this ‘parallel pierces the per- 
spective plane will be common to the traces of all the 
visual planes, and consequently, to the perspectives of. 
all the lines. 

74. The horizontal line drawn through the centre of 
the picture, is called the horizon of the picture, and will 
be the locus of the vanishing points of all horizontal 
lines. For, all horizontal lines drawn through the point 
of sight, are contained in the horizontal plane through 
the point of sight, and will therefore pierce the per- 
spective plane in the trace of this horizontal plane, 
which trace is the horizontal line drawn through the 
centre of the picture 

75. Horizontal lines, which make angles of 45° with 
the perspective plane, are called dzagonals.. If through 
the point of sight two diagonal lines be drawn, one on 
each side of the perpendicular to the perspective plane, 
the points in which they pierce the perspective plane 
are called the vanishing points of diagonals. These 
points are in the horizon of the picture, and at equal 
distances from the centre. 

76. Since the line which terminates siker of these 
points makes an angle of 45° with the perpendicular 
through the point of sight, as well as with the perspec- 


e 


104 TREATISE ON 


tive plane, it follows that the distance from the centre 
of the picture to the point of sight is equal to the dis- 
tance from the centre of the picture to the vanishing 
point of diagonals. 

Let AB (PI. 13. Fig. x) be the trace of the perspec- 
tive plane, which is perpendicular to the plane of the 
paper. 

Let C be the horizontal projection of the point of sight, 
and C’ its projection on the perspective plane: C’ is then 
the centre of the picture (70). Through the point of 
sight conceive two diagonal lines to be drawn—their 
horizontal projections are Ca and Cé, which make angles 
of 45° with the perpendicular CD. But these diagonals 
pierce the perspective plane at the points a’ and 6, in the 
horizontal line a’C’0, and these points, by definition, are 
the vanishing points of diagonals. Since the angles 
DCaand DCé are each 45°, Da and Dé are each equal to 
CD. But Ca’ and C’é' are each equal to Da or Dé, and 
consequently to DC, which is the distance of the point 
of sight from the centre of the picture. 

77. By considering what has been said, we have this 
general rule for determining whether the vanishing 
point of a diagonal is on the right or left of the 
centre of the picture. Through any point of the dia- 
gonal, in front of the perspective plane, draw a perpen- 
dicular to the perspective plane. Now, when the part 
of the diagonal intercepted between the point and the 
perspective plane, lies on the right of the perpendicular, 
the vanishing point of the diagonal is on the right of the 
centre of the picture; but when it les on the left of the 
perpendicular, the vanishing point is on the left of the 
centre of the picture. The rule is reversed when the 
point through which the perpendicular is drawn is behind 
the perspective plane. 


x 


LINEAR PERSPECTIVE. 10£ 


78. ‘The principles on which the perspectives of points 
are found, by the method we are now explaining, are 
these, 

1°. That the perspective of every point of a right 
line is found somewhere in the indefinite perspective of 
the line. 

2°. That the perspectives of two points of a right 
line determine the perspective of the line. From the 
first principle it follows, that if two lines be drawn 
through any point in space, and their perspectives de- 
termined, the intersection of their perspectives will be 
the perspective of their point of intersection. 

Let ¢ (Pl. 13, Fig. n) be a point in the horizontal 
plane at a distance ¢ d behind the perspective plane. 

The perpendicular through ¢ pierces the perspective 
plane atd; and since C’is the vanishing point of perpen- 
diculars, dC’ is the perspective of the perpendicular cd. 
The diagonal through ¢c pierces the perspective plane at 
f; and the diagonal has its vanishing point at a, there- 
fore fa’ is its perspective. But the perspective of the 
point ¢ is found both in dC’ and in fa’ ; itis therefore at 
c their point of intersection. 

We could have determined the perspective of the 
point ¢ by finding the perspectives of any other two lines 
passing through it; but it is better to use the perpen- 
dicular and diagonal than other limes, because their 
perspectives are more readily found. 

79. The perspective plane being placed between the 
eye and the object, the perspective of the object will 
lie above the ground line, and its horizontal projection 
will be behind the perspective plane. Now, when the hori- 
zontal plane is revolved in the usual way, to coincide 
with the perspective plane, the perspective, the horizon- 


106 TREATISE ON 


tal projection, and the vertical projection of the object 
will occupy the same part of the paper. 

To avoid this inconvenience in the constructions, we 
revolve the horizontal plane of projection about its inter- 
section with the perspective plane, in such a manner 
that the part behind the perspective plane shall_fall 
below the pe line. When this is done, the diagon- - 
als will have a direction in the construction contrary to 
their true direction in space. 

For example; had we so revolved the horizontal plane 
before finding the perspective of the point c, the horizon- 
tal projection of c, instead of beg above the ground 
line at c, would have been below it at g, dg being equal 
to de. In this case the diagonal gf would intersect the 
ground line at the point f, as before, and the perpendi- 
cular would meet the ground line at d. If then, we 
draw dC’ and fa’, we obtain ¢’, the perspective of the 
point as before. But there is this difference; were we 
to consider the point ¢ in front of the perspective plane, 
which we ought to do if the horizontal plane were re- 
volved in the usual way, the diagonal g¢ f would have its 
vanishing point at 6’, and the perspective of the point 
would bebelow the ground line. 

We may then project points which are behind the per- 
spective plane, on the part of the paper in front of the 
sround line, by observing that the true vanishing point 
of any diagonal is the one opposite to that which its 
projection indicates. 

80. If through the point of sight a plane be passed 
parallel to any plane in space, it will contain all lines 
drawn through the point of sight and parallel to the lat- 
ter plane. Hence, the trace of the visual plane will be 
the locus of the vanishing points of all lines contained 
in the parallel plane. 


a ae 


rN 


LINEAR PERSPECTIVE. 107 


Therefore, the vanishing line of any plane is the trace, 
on the perspective plane, of the visual plane drawn 
parallel to it. 


PROBLEM ff. 


Having four cubes placed in the four angles of a square, 
their bases in the same horizontal plane, it is required to find 
the perspective of the cubes and the perspective of their shad- 
ows on the plane of their bases. 


81. [f the perspective plane were taken through the 
front faces of the cubes whose horizontal projections 
are cr and ds (Pl. 13, Fig. m), AC would be its horizon- 
tal trace, and the front faces of the cubes being in the 
perspective plane, would be their own perspectives. 
The projection in this figure, although made ona small 
scale, shows the position which the cubes have with 
each other, and with the perspective plane. 

Let us also suppose the point of sight to be in a plane 
equidistant from the inner faces of the cubes. 

Draw any line, as CD, for the ground line of the plane 
on which the perspective is to be made. 

Assume E for the centre of the picture, draw the ho- 
rizontal line H’EH, and take H and H' for the vanishing 
points of diagonals. Then EH, or EH’ is equal to the 
distance of the point of sight from the perspective 
plane. 

Draw EE’ perpendicular to the ground line, and 
from E’ lay off E’d and E’c, each equal to ‘half the dis 
tance between the inner faces of the cubes. Make dé 
and ca each equal to the length of an edge of the given 
cubes, and on them coristruct the squares d6' and ca; 
these squares are the perspectives of the faces which 


108 TREATISE ON 


are in the perspective plane. Through the points a c, 
d, b, a’, c,d’ and &’, draw lines to E, the centre of the pic- 
ture. ‘These lines are the indefinite perspectives of the 
lines ag, cp, dl, bh (Fig. m), and of the parallel edges 
directly over them (72). 

Through a and 6 draw the lines aH and 6H’ to the 
vanishing points of diagonals; these lines are the indefi- 
nite perspectives of ah and bq, Fig. m. 

The points e, m, f and g, in which the perspectives of 
the diagonals intersect the perspectives of the perpen- 
diculars, are, respectively, the perspective of the points 
é, m, f, and g, Fig. m. 

Through e draw e¢ parallel to ce’; the point é, in 
which it intersects ci, is the perspective of the angular 
point of the cube which is horizontally projected at e, 
Fig.m. For the edge of the cube which pierces the hori- 
zontal plane at ¢ is parallel to the edge which pierces it 
at c, and both are parallel to the perspective plane; 
hence their perspectives are parallel (73). But both 
these edges are limited by the edge which pierces the 
perspective plane at c’, and the indefinite perspective of 
this latter edge is cli; therefore ¢ is the perspective of 
the angular point directly over the one whose perspective 
ise. Similar reasoning will apply to the other vertical 
edges of the cubes. 

Through e draw er parallel to ac; the point r, where it 
meets ak is the perspective of the angular point r, Fig. 
m. From yr drawrr’ parallel to aa’, and from the point 
r where it intersects aE draw r'é. 

We have then determined the square acca’, the per- 
spective of the face in the perspective plane: the trape- 
zoid acer, the perspective of the base of the cube: the 
trapezoid cecc’, the perspective of a face perpendicular 
to the perspective plane: the square ree’, the perspec- 


LINEAR PERSPECTIVE. 109 
Ave of the face parallel to the perspective plane; the 
trapezoid ¢e’r'a’, the perspective of the upper face of the 
cube: and the trapezoid arr‘a’, the perspective of a 
second face perpendicular to the perspective plane. 

The perspective of the angular point m of the cube mq, 
Fig. m, has already been determined; and since the edge 
mm, Fig. m, is parallel to the perspective plane, the per- 
spective of the point x must lie in mn, drawn through m, 
parallel to the ground line; but it is also in aH, the in- 
definite perspective of ag; hence it is at n»their pomt 
of intersection. 

Through m and n draw vertical lines and produce 
them till they meet the lines ¢Eand a’E, and jointhe points 
of intersection; we have then the perspective of the 
face parallel to the perspective plane. 

Throughn drawnH to the vanishing point of diagonals: 
this line is the perspective of np, Fig. m, and the point p, 
where it intersects cli, is the perspective of the point p, 
Fig.m. Through p draw pq parallel to the ground line 
till it meets aE; and at g and p erect perpendiculars to 
the ground line, and complete the perspective of the 
cube as in the last case. The perspectives of the other 
cubes are determined in a manner entirely similar. 

It remains to find the perspective of the shadows cast 
on the horizontal plane. 

82. The shadow which a point casts upon a plane, is 
always found in the ray of light passing through the 
point, and also in the projection of this ray upon the 
plane on which the shadow falls. Hence, the perspec- 
tive of the shadow will be found in the perspective of the 
ray, and in the perspective of its projection, and is con- 
sequently their point of intersection. | 

83. ‘The perspective of the shadow cast by a right line 


110 TREATISE ON 


on a plane, is in the indefinite perspective of the inter 
section of a plane of rays passed through the line, with 
the plane on which the shadow falls. 

64. Since the rays of light are parallel, they have a 
common vanishing point (71). This point is where a 
ray of light drawn through the point of sight pierces the 
perspective plane. 

Let (E, E’), Fig. 2, be the direction of a ray of light. 
Through the point of sight (C, C’) let a ray of light be 
drawn, it will pierce the perspective plane at R, which 1s, 
therefore, the vanishing point of rays. 

The projections of rays on any plane are also parallel 
to each other, and consequently they have a common 
vanishing point. 

Through the point of sight let a line be drawn parallel 
to E, the horizontal projection of a ray of light. This 
parallel pierces the perspective plane at I’, which is, 
therefore, the vanishing point of the horizontal projec- 
tions of rays. 

Now, since the vertical plane of rays through the 
point of sight contains the ray of light through the 
point of sight, and also the line drawn parallel to the 
horizontal projections of rays, it follows, that its trace 
on the perspective plane will contain both the vanishing 
point of rays and the vanishing point of horizontal pro- 
jections. But its trace on the perspective plane is a 
vertical line; hence, the line joining the vanishing point of 
rays and the vanishing point of horizontal projections, is per- 
pendicular to the ground line. 

85. We see that the vanishing point of rays and the 
vanishing point of horizontal projections can be found 
when we know the direction of the light. Reciprocally, 
if we know the vanishing point of rays and the vanish- 


LINEAR PERSPECTIVE. 11) 


ing point of horizontal projections, we can determine 
the direction of the light. 

For, the line joining the vanishing point of rays and 
the centre of the picture, will be parallel to the projec- 
tions of the rays on the perspective plane. And if 
through the centre of the picture, the perpendicular C’D 
be drawn to the ground line, and produced, and DC 
made equal to the distance from the centre of the pic- 
ture to the vanishing point of diagonals, we shall have 
C the projection of the point of sight on the horizontal 
plane. Joiming this point with I", the point in which the 
line joining the vanishing point of rays and the vanishing 
point of horizontal projections intersects the ground line, 
and we have Cf” the horizontal projection of the ray of 
light passing through the point of sight. 

86. In finding the shadows cast by the cubes on the 
horizontal plane, let Rh be the vanishing point of rays, 
and P the vanishing poit of horizontal projections of 
rays. 

By considering the direction of the light, it is plain, 
that the edge of the cube whose perspective is ec’ will 
cast a shadow on the horizontal plane. But the plane 
of rays through this edge is perpendicular to the hori- 
zontal plane; Tetoe: the shadow of the edge is parallel 
to the horizontal projection of the rays, and consequently 
P is its vanishing point. But the shadow is limited by 
the ray through the point whose perspective is c’; there- 
fore, its perspective is limited by the perspective of this 
ray, whichis ¢R: consequently, c¢ is the perspective of 
the shadow. 

The next line which casts a shadow on the horizontal 
plane is the edge whose perspective is ce’. Since this 
line is perpendicular to the perspective plane, it will be 
parallel to the horizontal plane, and therefore its shadow, 


vey 


ee ee | ” no fe 
k . SL Ae she 


* 


132 TREATISE ON 


which is parallel to itself, will be perpendicular to the 
perspective plane; and hence, its vanishing point is at 
EK. But tis one point of its perspective; therefore ¢E is 
its indefinite perspective. This perspective is limited 
by é¢R drawn to the vanishing point of rays. 

The shadow cast by the edge whose perspective is 
r'¢ is parallel to the line itself, and also to the perspec- 
tive plane; hence its indefinite perspective is nv, drawn 
through n parallel to the ground line. This perspective 
is limitedby rR, and also by rP. The line rv is the per- 
spective of the shadow cast by the vertical edge of the 
cube whose perspective is r7’.. Only a small part of this 
shadow is seen, nearly all of it being behind the cube. 

The shadows of the other cubes are found ina manner 
so entirely similar, as not to require Meee dL expla- 
nations. 

The faces of the cubes which are in the shade are 
darkened in the perspective. 


*, PROBLEM MUI. } 


Having given four pyramids, standing on pedestals, and 
situated in the four angles of a square, it ts required to find 
the perspectives of the pyramids and ROSES and the per- 
spectives of their shadows. 


87. Fig. n (Pl. 14) represents, on a small scale, the 
projections of the four pyramids and pedestals, having 
the same relative position as those which are to be put 
in perspective. 

Let the perspective plane be taken through the front 
~ faces of the pedestals: AB is its horizontal trace. 

From any point, as a, lay off a5 equal to the side of 


‘+ rete ww 


the square which forms the base of the pedestal, and on 


© tg 


; . 
bor, te 


oa * 


aa 
ay 
LJ 


S 
17 


—_ 


Plate 15 
a . — - ~ g z " " — 3 ~ = _ - _ ™ ewe - oe -_ - " = oe ~- - * = 


S o y ( 
. > fy ' 
Jen ij a 


2 ee ——s 
eee ee a eee B r | 
== : Hh 
Z = : : | 
3 ae | = 
igen. aN Kes a as ae a 
CO | we 
= SS R 
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| : 
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C a Cc 


Prudhomme Sc. | 


# 
LINEAR PERSPECTIVE. 113 


it describe a rectangle equal to the front face of the 
pedestal. From 6 lay off dc equal to the distance be- 
tween the pedestals, and make cd equal toad; and oncd 
describe a rectangle equal to the front faces of the 
pedestals. ‘The perspectives of the four pedestals are 
then determined by constructions entirely sna to those 
of the last problem. 

To find the perspectives of the vertices of fre pyra- 
mids : 

Let S be the centre of the picture, and D and D the 
vanishing points of diagonals. 

{f a line be. drawn through the vertices of the two 
pyramids on the left, it will be perpendicular to the per- 
spective plane, and will pierce it in the vertical line AX, 
drawn through the middle point of a. On this line, 
therefore, lay off hh’ equal to the height of the vertices 
of the pyramids above the base of the pedestals, and /’ 
will be the point in which the line through the vertices 
of the pyramids pierces the perspective plane. The 
line /'S is the perspective of this perpendicular. 

The diagonal through the vertex of the front pyramid 
pierces the perspective plane in the horizontal line h’a, 
and also in the vertical line aa’; hence aD" is the per- 
spective of the diagonal, and v the perspective of the 
‘ertex of the front pyramid. The perspectives of the 
vertices of the other pyramids are easily found. 

From e lay off ef equal to the distance which the 
pedestal projects beyond the pyramid, draw eD’ to the 
vanishing point of diagonals, and from f draw /S to the 
vanishing point of perpendiculars ; ; the point g, in which 
these lines intersect, is the perspective of the point in 
which an edge of the pyramid pierces the upper face of 
the pedestal, and the horizontal line gp 1s the indefinite 


PIO 
ere 


om 


114 TREATISE ON e 


perspective of the line in which the front face of the 
pyramid intersects the upper face of the pedestal. 

From z lay off on ce a distance equal to e f, and 
through the point so determined draw a line to the centre 
of the picture : the point p, in which it meets gy, is the 
-perspect awe of the point in which a second ae of the 

pyramid pierces the upper face of the Redentall The 
diagonals through p and ¢ determine, by their intersec- 
tions with the eden gS and pS, the points in 
which the two remaining edges pierce the upper face of 
the pedestal. Joining these points with v, the perspec- 
tive of the vertex, we have the perspective of the pyramid. 

Only a part of the edge of the pedestal which is per- 
pendicular to the perspective plane at e,is seen. ‘The 
perspective of the edge intersects the perspective of the 
edge of the pyramid at / If through the poimt/a 
visual ray be drawn, it will intersect, in space, both the 
edge of the pyramid and the edge of the pedestal. At 
the point in which it intersects the edge of the pedestal, 
the edge of the pedestal passes behind the pyramid, and 
is not seen; and the same may be said of the edge of 
the pedestal parallel to ez. The perspectives of the 
other pedestals and pyramids are found by constructions 
entirely similar. 

To find the perspectives of the shadows cast on the 
horizontal plane: 

Let R be the vanishing point of rays, and P the 
vanishing point of horizontal projections. 

The point v’, in which the diagonals pD and gD’ 
intersect, is the perspective of the projection of the 
vertex of the pyramid on the plane of the upper base of 
the pedestal; hence, v’P is the perspective of the pro 
jection of the ray through the vertex of the pyramid, on 
that plane. But vR is the perspective of the ray; 


LINEAR PERSPECTIVE. 115 


nence ¥" is the perspective of the shadow cast by the 
vertex of the pyramid on the plane of the upper face 
of the pedestal. 

It is plain that the edges of the pyramid which pierce 
the upper face of the pedestal in the points whose per- 
spective are p and &, will cast shadows on the pedestal 
and on the horizontal plane. Therefore, pv’ and kv’ are 
the indefinite shadows on the upper face of the pedestal. 
It is evident, that only the parts pm and kn fall on the 
pedestal, and that the points m and 2 cast shadows on 
the horizontal plane. 

The point 7, in which the diagonals aD’ and 6D inter- 
sect, is the perspective of the projection of the vertex of 
the pyramid on the plane of the base of the pedestal. 
From r, draw rP to the vanishing point of projections, 
and the point v”, in which it intersects vR, is the per- 
spective of the shadow cast by the vertex of the pyramid 
on the horizontal plane. 

The line 62’ is the perspective of the shadow cast on 
the horizontal plane by the edge 52 of the pedestal, and 
aS is the indefinite perspective of the shadow cast by zs. 
Therefore, drawing through m, mR to the vanishing point 
of rays, determines m’, the shadow cast on the horizontal 
plane by the point m: the shadow z’m‘ is cast by ¢ m. 

The part ms, of the edge zs, will not cast a shadow on 
the horizontal plane, being itself in the shadow of the 
pyramid. If, however, we draw from s a line to the 
vanishing point of rays, the point in which it mtersects 
2'S, limits the shadow which would fall on the horizontal 
plane if the edge were in the light. The line drawn 
through s’, the point so determined, and parallel to the 


ground line, is the indefinite perspective of the shadow | 


cast by the edge Xs. Through » draw »R, and we de- 


termine n’, the shadow cast on the horizontal plane by 
H 2 


a 


#. 


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116 TREATISE. ON a 


. the point n; and n'v” is the shadow cast by the edge of 
the pyramid. The part ns, of the edge of the pedestal, 
being in the. shadow of the pyramid, cannot cast a 
shadow on the horizontal plane ; ang the perspective of — .. 

. . the shadow cast by Xn begins at n’,and is paralleltothe - 

* ground line. * 
' ‘The perspectives of the shadows of the other pyra- | 
‘ mids are found by similar constructions. ‘The faces of 
¥* the.pyramids and pedestals which are in the shade and 
; : seen, are shaded in the drawing. 
+ 


PROPOSITION IV. THEOREM. . 


If aright line be tangent to a curve in space, the perspective E;’. 
i . ‘of the right line will be 1 tangent to the perspective of the curve. 


88. For let AFCG (PI. 15, Fig.1) bea curve, to which 


a right he is dra awn tangent at any point as F. 


and a visual Satiy through the right line, the visual — 
= plane will be tangent to es visual cone. ‘T’he perspec- 
_ tive plane will intersect the visual plane in a right line 
a and the visual cone in a curve, and the right line and | 
; curve will be tangent to each other (Des. Geom. 84). 
, But the right line in which the perspective plane inter- 
» sects the visual plane is the perspective of the tangent ~ 
line,.and the curve in which the perspective plane inter- 
sects the visual cone is the perspective of the given curve 
AF CG: hence, when a right line and curve are tangent * 
‘In space, their perspectives are also tangent. 
. If two curves are tangent to each other in space, their 
. perspectives are also tangent. For the two visual — 
a cones which determine their perspectives are tangent . 
y to each other, and therefore the curves in which they ~ 
Bes intersect the perspective plane are likewise tangest.  - 


bt a 
wt 
* 


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\LINEAR PERSPECTIVE. ’ MWh ee 
2 


* 
* 
PROBLEM V. 


To find the perspecige of a aie. , 


89. Let B (PII arty: be the centre of the circle 
which is to be put active and RT the trace of 
the perspective plane; the perspective plane being per- 
pendicular to the plane of the circle AFCG. 

Althotgh the horizontal projection of the circle is 
made in front of the perspective plane, all the points of 
it are, in fact, as far behind it as they are now projected ? 
in front of the trace RT. : 

Let the point of sight be taken in a plane passing 
through the centre of the circle, and perpendicular to 
the perspective and horizontal planes. . 

Let S be the centre of the picture, and P and P x 
the vanishing points of diagonals. Through B draw, 
the diagonal BN, and ae a perpendicular to the | 
ground line. Oe 

If the circle were in front of the perspective plane, 7 x 
the diagonal BN would haye its vanishing point at P;  ..’ 
but since it is behind it, ‘the vanishing point. of the ~ 
diagonal is at P’ (79). Hence NP’ is the perspective. 
of the diagonal: The perpendicular through B piérces . a 
the perspective plane at n, and has its vanishing point » 
at S; therefore the point 6 where Sn intersects the per- > me 
spective of the diagonal, is the perspective of the iat 
centre B. 5 a 

Through 6 draw the horizontal line «4 c} this line ys 
is the urate perspective of the diameter ABC. | © 
Through A and C draw tangent lines. These tangents 
are perpendicular to the ground line, and their perspec- 
tives pass through 8. . me pone a and c in which 


a 
» 
4 
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> s co,  * ~~ 
. a « * ” 
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ve I 


in : 


ee 


138 . TREATISE ON 


they intersect the perspective of AC) are the perspectives 
of the points A and C. The lines Sa and Sc are tan- 
gent to the ellipse, which is the perspective of the circle 
AFCG. 

Let the perspectives of the points F and G be next 
found; they are / and g. fpr a 

If through the points F and G tangent lines be drawn 
to the circle AFCG, their perspectives will be tangent 
to the perspective of the circle (88). But since the 
tangents are parallel to the ground line, their perspec- 
tives will also be parallel to the ground line; hence gf 
is perpendicular to the tangents drawn through its ex- 
tremities f and g; it is, therefore, an axis of the ellipse, 
which is the perspective of the circle. 

Bisect gf atd. ‘Throughd draw Pde, and from e draw 
the diagonal eD. It is plain that Pde is the perspective 
of the diagonal De, and that dis the perspective of D. 

Through D draw KDH parallel to the ground line, 
and find the perspectives of the points K and H, which 
are & andh; kh is the: perspective of KH, and is the 
other axis of the ellipse. ‘The ellipse therefore can be 
described. 
~ 90. When the point of sight is not in the plane pass- 
ing through the centre of the given circle and perpen- 
dicular to the perspective and horizontal planes, we are 
unable to find the axes of the ellipse by a direct construc- 
tion. We then find the perspectives of several points 
of the circumference of the circle, and describe the 
ellipse through them. 

In Pl. 15, Fig. 2, the perspective of the circle is found 
by points. The perspectives of the tangents at, the 
points a, d, 6, g,c and e are tangent to the perspective 
of the circle at the points a’, d’, 6, g’, ¢ and é. 

91. Having given a circle in space, and the point of 


“ 


es 


LINEAR PERSPECTIVE. 119 


sight, we may so place the perspective plane that the 
perspective of the circle shall be any.one of the conic 
sections. 

For, when the circle and point of sight are given, the 
visual cone circumscribing the circle is also given, and 
if the position ofthe perspective plane be undetermined, 
it may be so chosen as to intersect the cone in any 
one of the conic sections. 

When the perspective plane is parallel to the base of 
the visual cone, or when it cuts the cone in a sub-con- 
trary section, the curve of intersection is a circle. 


Me PROBLEM VI. 


To find the perspective of a cylinder, the perspective of the 
shadow cast by the upper circle on the interior surface, and 
the perspective of the shadow on the horizontal plane. 

re 


92. Let the circle described in the horizontal plane 
with the centre C (PI. 15, Fig. 3), and radius CB, be the 
lower base of the cylinder; the centre C being ata dis- 
tance behind the perspective plane equaltoCC”. Let A’B’ 
be the projection on the perspective plane of the upper 
base of the cylinder, the plane of this base intersecting 
the perspective plane in the horizontal line A’B’. 

Let the point of sight be taken in the plane through 
the axis of the cylinder and perpendicular to the per- 
spective plane. Let S be the centre of the picture, and 
P’ and P the vanishing points of diagonals. 

Find now the perspective of the lower base of the 
cylinder as in Prob. 5. 

In finding the perspective of the upper base we have 
merely to regard the perpendiculars and diagonals 
already drawn, as the projections on the horizontal plane 


» 
4 


120 TREATISE ON 


of corresponding perpendiculars and diagonals drawn in 
the upper base of the cylinder. 

For example, the diagonal aBé, being considered in 
the upper base of the cylinder, would pierce the perspec- 
tive plane at 0’, its perspective would be a’P’, and its 
intersection with C’S determines a’, a point in the per- 
spective of the upper base; the diagonal Be determines 
the point c’. 

If through the point of sight two tangent planes were 
drawn to the cylinder, they would touch it in the two 
elements which pierce the horizontal plane at fand g. 

Having found g’ and/’, the perspectives of the points 
g and f, in the lower base of the cylinder, and g” and f” 
the perspectives of the corresponding points of the upper 
base, draw the lines g’g" and f’f"; these lines are the per- 
spectives of the elements which pierce the horizontal 
plane at g andf- 

The part of the cylinder convex towards the point 
of sight, and limited by these elements, is seen; the 
other part is not seen. Therefore, the semi-ellipses g of 
and g"c'f", which are seen, are made full, and the semi- 
ellipses g’af’ and g’a'f", which are not seen, are dotted. 

To find the Den Kosai of the shadow cast on the 
interior of the cylinder by the circumference of the 
upper base: 

Let R be the vanishing point of rays, and H the 
vanishing point of horizontal projections. 

If two tangent planes of rays be drawn to the cylin- 
der in space, their horizontal traces will be tangent to 
the base of the cylinder, and the elements of contact 
will be the elements of shade. But the horizontal traces 
of these planes will be parallel to the horizontal projec- 
tion of the rays of light; hence, their vanishing point 
is at H. The horizontal traces are also tangent to the 


am," 
LINEAR PERSPECTIVE. 121 


base of the cylinder, therefore their perspectives will be 
tangent to the perspective of the base (88). 

Through H draw the tangents Hk and Hh; the points 
of contact &# and A are the perspectives of the two 
points in which the elements of shade pierce the hori- 
zontal plane. But since the elements of shade are ver- 
tical lines, kX’ and hh, drawn perpendicular to the 
ground line, are their perspectives, and #’ and fh’ are the 
perspectives of the points at which the shadow on the 
interior of the cylinder begins. 

if we suppose the cylinder in space to be intersected 
by a plane of rays parallel to its axis, the horizontal 
trace of the plane will be parallel to the horizontal pro- 
jection of the rays of light, and consequently, will have 
its vanishing point at H. Every plane so drawn will in- 
tersect the cylinder in two elements, and the one towards 
the source of light will cast a shadow on the other. 

hrough H draw any line, as Haz, to represent the 
perspective of the horizontal trace of a secant plane of 
rays. Through the points z and m, in which it itersects 
the perspective of the base, draw the elements 77’ and 
nn. Through? draw? R to the vanishing point of rays ; 
the point m in which it intersects nn’ is the perspective 
of a point of shadow on the interior of the cylinder. 

To find the shadow cast on any particular element, as 
ff’, draw from the vanishing point of horizontal pro- 
jections a line, as Hf’, through its foot, and through the 
upper extremity of the element passing through the other 
point in which Hf’ intersects the perspective of the base, 
let a line be drawn to the vanishing of rays; the point 
p, in which it intersects the element /"/", 1s the point of 
shadow required. 

To find the perspective of the shadow cast on the 
horizontal plane : 4? 


» & 


z ’ ~ : *% | = 
> 4 : 

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3 rs . *. s :” 
‘ ‘te . mm r 
: | oo 

”~ , > « 

t 222 . TREATISE ON 


The traces of the tangent planes of rays are tangent 
to the shadow cast by the upper circle of the cylinder 
on the horizontal plane (27); therefore, their perspec- 
tives are tangent to the perspective of that shadow (88). 
But the rays of light passing through the upper extremi- 
ties of the elements of shade, intersect the traces of 
the tangent planes at the points of tangency (27); 
therefore, /” and h’, where the perspectives of the rays 
intersect the perspectives of the traces, are the perspec- 
tives of the points of tangency, and hh’, kk” are the per- 
spectives of the shadows cast by the elements of shade. 

That part of the upper base whose perspective is 
kan fh’, casts the curve of shadow on the horizontal 


*- 


of the cylinder. | 
Any line, as Hz, drawn through H, may be considered 


as the perspective of the horizontal trace of a plane 


of rays; the point n”, in which the perspective of the ray 
» through n’ intersects Ha, is the perspective of a point 
of shadow on the horizontal plane. 
A line drawn through R, tangent to the perspective 
of the upper» base, will also be tangent to the curve 
ey aa + so 
leigh. te Me “s 
The part. of the surface of the cylinder which is in 


al , "the shade, and seen, is shaded in the drawing. The 


perspective Of the shadow on the horizontal plane is 
_ also shaded. : 


aa * 
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a 
t 
a 4 
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plane ; the remaining part casts a shadow on the interior 


Lee a DS 


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LINEAR PERSPECTIVE. @ 123 
e 
PROBLEM VIL » oo” 


It ts required to find the perspective of the frustum of an 
inverted cone ; also the perspective of the shadow on the ante- 


rior of the frustum, and the perspective of the shadow ow the 


horizontal plane. * 


93. Let the circle described with the centre A and 


tadius AB (Pl. 16), be the horizontal projection of the 
upper base of the frustum, and B’CG’ the intersection 
of its plane with the perspective plane. Let the circle 
described with the centre A and radius AH, be the lower 
base of the frustum, and HT its vertical projection. 
Lhe horizontal projection of the vertex of the cone 


is at A, and its vertical projection, which is at L, is found » 


by joining B’ and H’-and producing the line until it in- 
tersects CAE drawn perpendicular to the ground dine, 
Let S be the centre of the Shes and D the vanish- 
ing point of diagonals. ‘ 
; ‘Through (E, C), agpomt of the ee base, draw a 
diagonal and perpendicular. The diagonalpierces the 
perspective plane at G’, and the perpendicular pierces it 
at C. The diagonal’ has its vanishing pointrat*D, and 
the perpendiculaz its vanishing. point at S. 
e is the perspective of the point (E,C). By similar con- 
structions we find ‘, the perspective of. (H,C), ¢ 


the perspective of the point whose horizontal projec- - 


tion is c, and 6 the perspective of (B,B'). The perspec- 
tive of any point may be found by determining the per- 
spectives of the diagonal and perpendicular passing 
through it. 

Before describing the ellipse bec’, it will be. well to 
remember that ¢S and B'S, being the perspectives of 


Therefore, - 


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124 - TREATISE ON - 


- 


tangent lines to the upper base of the frustum, are tan- 
gent to the ellipse deceé : and also, that the tangent lines 


In space drawn through the points (E,C) and (E’,C) are 


parallel to the perspective plane; hence, their perspec- 
tives are the herizontal lines drawn through e and é, 
tangent to the ellipse deceé. 

The perspective of the lower base of the frustum is 
determined by constructions entirely similar. 

Let the perspective of the vertex of the cone be next 
found. ‘i 

The ornate through the vertex of the cone 
pierces the perspective plane at L, the vertical projec- 
tion of the vertex; and the diagonal through the vertex 
pierces the perspective plane at N’; therefore the point 


L’, where LS intersects N’D, is the perspective of the - 


vertex. 

if through the point of sight we suppose two planes 
to be drawn tangent to the frustum of the cone, the 
traces of these planes, on the perspective plane, will 
pass through the perspective of the vertex, and will 


limit the perspective of the cones hence, the perspec- 


tives of the upper and lower circle will be tangent to 


these traces. Let these tangents be then drawn through | 
the point'L’. 
The point of sight being above the upper base of the - 


frustum, the whole of that circle will be seen, and there- 
fore its perspective is made full. A part only of the 


‘ lower circle is seen, the perspective of this part is 


made full, and is limited by the tangent lmes drawn, 
through L’. 

To find the shadow which the upper ce casts on 
the interior of the frustum : 

Let R be the vanishing point of rays, and P the 
vanishing point of horizontal projections. 


_ LINEAR PERSPECTIVE. 125 


Since the ray of light through the vertex of the cone 
is a line of every plane of rays which intersects the cone 


in right-lined elements, the point in which this -ray 


pierces the horizontal plane is common to the horizon- 
tal traces of all such secant planes; hence, the perspec- 
tive of this point is common to the. perspectives of all 
the traces. 

But the perspective of this point is found in the per- 
spective of the ray through the vertex of the cone, and 
in the perspective of the horizontal projection of this 
ray (82). Through R draw RU’; this line is the indefi- 
nite perspective of the ray. Through a’, the. perspec- 
tive of A, draw Pa’; this is the indefinite perspective of 
the horizontal projection of the ray; the point K, in 
-which they intersect, 1s the perspective of the point in 
which the ray through the vertex of the cone pierces 
. the horizontal plane. 

Through K, draw Kf and Kd tangent to the perspec- 
tive of the lower base of the frustum. These tangents 
are the perspectives of the horizontal traces of the two 
planes of rays which are tangent to the frustum in space. 
Hence, L/h and Lidg are the perspectives of the elements 
_of shade, and g and / the perspectives of the points at 
which the shadow on the interior of the frustum begins. 
To find points of this shadow, draw any line through 
K, as Kpqs', which will be the perspective of the hori- 
zontal trace of a secant plane of-rays. Through the 
points p and gq draw the elements L’pk and L’gs. From 
k draw /R to the vanishing point of rays; the point # 


in which it intersects L’s, is the perspective of a point . 


of shadow. The shadow on any particular element is 
found by drawing a line from K through its foot; and 
then drawing the perspective of aray through the upper 


a” 


. 


ts 3 ‘ ey TREATISE. on Ly od mata ee + 
“He ofthe chy ti ht, as 1 

efor i va ra 
ats The part, of F the curve. whose Pe is sabhdh 
bY Ms , casts. a shadow. on the interior of the frustum ; and the .: 
at part. whose perspective. jae seg casts a hago g on aA 
how alplave, <¢.!* osha 
- Ttis is now oe ind tg shadow on the ho horizontal 3 

arte ate | ae ; | ‘A 
) A Bice Pod ‘point | whose pbaspeatine is h, casts a tate: in x A 
ca thé trace ‘of the tangent lane of rays; but the per- Ae. 
.. spective of the shadow i 18 so in the perspective of the a me 
ray through h; hence it is at h’.. Fora similar reason. bP ae 


Soe 


ey g’ is the perspective of a point of the shadow, on ‘the 
ae horizontal plane.» * ae ae Ye x 
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. : 
, PROBLEM VIII. 


To find the perspective of aniche, and the per eee ms the 


shadow cast on tts interior surface. 


94. Let the perspective plane be taken through the 
front face of the niche. Let AB (PI. 17) be the ground 
line, and the semicircle 6ar the horizontal projection of 
the niche. * 

_ Draw 4c and re, in the perspective plane, perpen 
dicular to the ground line, and make them equal to the 
height of the cylindrical part of the niche; and on ce 
describe the semicircle cle. ‘These are the lines of the 
_ niche which are in the perspective plane. 
_ The perspective of the lower base of the niche is 
- rahb, the segment of an ellipse: the perspective of the 
semicircle of contact of the cylindrical and spherical 
parts, is the elliptical segment extpe; both of these curves 
are found by methods already explained. The lines 6S 
“and 7S, are the perspectives of the two lines drawn tan- 
gent to the base of the niche at the points 6 and 7; hence 
they are tangent to the elliptical segment ra‘hd. (88.) 

To find the shadow on the interior of the niche. The 
first line which casts a shadow on the interior of the 
niche, is the element c 6. If through this element a plane 
of rays be passed, it will intersect the surface of the 
niche, in space, in a second element on which the shadow 
will fall. But, since the plane of rays is vertical, its 
horizontal trace is parallel to the horizontal projection 
of the rays of light; hence its vanishing point is at H, 
the vanishing point of horizontal projections of rays, 
and 6H is its perspective.’ 


125 TREATISE ON 


Through d, the point in which 6H intersects ra’b, draw 
dd'paralleltodc; dd’ isthe indefinite perspective of the 
element that receives the shadow. ‘Through ce, the upper 
extremity of the element casting the shadow, draw cR 
to the vanishing point of rays; the point d’, in whichat. 
intersects dd’, is the perspective of the shadow cast by 
the point c, and dd’ is the shadow cast on the cylindrical 
part of the niche, by the element bc. The line éd is the 
perspective of the shadow which a part of the same 
element casts on the horizontal plane. 

‘Through H draw any line, as Hf, near to Hd. This 
line may be regarded as the perspective of the horizontal 
trace of a vertical plane of rays; andAf'is the per- 
spective of the element in which it intersects the cylin- 
drical part of the niche. The plane intersects the per- 
spective plane in the line ff’; therefore the point A’, in 
which /’R intersects hi’, is the perspective of the shadow 
cast by the point (7, f’) onthe cylindrical part of the niche 

To find the perspective of the shadow which falls on 
the spherical part of the niche: 

If we suppose the quadrant of the sphere, which 
forms the spherical part of the niche, to be intersected 
by a plane parallel to the front face of the niche, the 
section will be a semicircle whose diameter will be a 
chord of the semicircle of contact of the cylindrical 
and spherical parts of the niche. The front circle of 
the niche will cast a shadow on this plane equal to itself 
(28); and the point where this shadow intersects the 
circle cut from the sphere is a point of the required 
shadow in space. If then, we find the perspectives of 
these two circles, the point in which they intersect will 
be the perspective of a point of the curve of shadow. 

Since both the circles are parallel to the perspective 
plane, their perspectives will be circles (91). Draw any 


od 


LINEAR PERSPECTIVE. 4 129 


line, as np, to represent the perspective of the diameter 
of the semicircle cut from the spherical part of the niche 
by the parallel plane, and on it describe the semicircle 
nkp. ‘he perspective of the shadow cast on the plane 
__ of this circle by the centre g, is found in gR, and also 
in the perspective of the projection of the ray through 
g on that plane. But, since the plane is parallel to the 
perspective plane, the projection of the ray upon it is 
parallel to SR. But g’ is the perspective of one point 
of its projection; hence g‘g”, drawn parallel to SR, is the 
perspective of the projection of the ray on the parallel 
plane, and g” is the perspective of the shadow cast by 
the centre g. The shadow cast by cg is parallel to itself 
and to the perspective plane; hence ¢”g, drawn parallel 
to cg, 1s the perspective of the indefinite shadow cast by 
eg on the parallel plane. But this shadow is limited by 
the ray cR; hence g’g is the perspective of the radius of 
the circle of shadow. With yg” as a centre, and the 
radius gg, describe the are gk ; the pomt £, in which it 
intersects xkp, is the perspective of a point of the curve | 
of shadow. 

If through the centre g, gt be drawn in the perspec- 
tive plane, and perpendicular to SR, it will be a line of 
the plane of the circle of shadow; therefore the point 2, 
in which it meets the circumference c / e, is a point of 
the perspective of the curve of shadow. 

We can easily find the perspective of the point at 
which the shadow passes from the spherical to the 
cylindrical part of the niche. For, if through the point 
in space of which & is the perspective, a line be sup- 
posed drawn parallel to gz, it will be contained in the 
plane of shadow, and will pierce the upper base of the 
cylinder in the trace of the plane of shadow. But ks, 
drawn parallel to gz, is ng perspective of this parallel, 


130 TREATISE ON 


and s is the perspective of the point in which it pierces © 
the upper base of the cylinder. Hence, gst is the per- 
spective of the trace of the plane of shadow, and ¢ the 

perspective of the point at which the shadow passes — 
from the spherical to the cylindrical surface. 3 

There are other constructions for finding the shadow 
on the spherical part of the niche. 

Revolve the plane of rays. passing through the 
point of sight and perpendicular to the perspective 
plane, about SR, until it coincides with the perspective 
plane. The point of sight falls in SD”, drawn perpen- 
dicular to SR, and at a distance from S equal to SD 
(76). The ray of light through the point of sight takes 
the position RD”. 

Intersect the spherical part of the niche by a plane of 
rays perpendicular to the vertical plane. Let Im’z, 
drawn perpendicular to gz, or parallel to SR, be the 
trace of such a plane. Revolve this plane until it coin- 
cides with the perspective plane. The semicircle cut 
out of the sphere, when revolved, is the semicircle lmz, 
and the ray through / takes the position /m, parallel to 
RD”. Let the counter revolution be now made, and 
draw mS, which is the perspective of mm’, and JR, which 
is the perspective of the ray; the point m’, in which they 
intersect, is the perspective of a point of the curve of 
shadow. Other points may be found by similar con- _ 


structions. 


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LINEAR PERSPECTIVE. P 131 


PROBLEM IX. 


To jind the perspective of a sphere, the perspective of tts 
shade, and the perspective 8 tts shadow on a horizontal 


plane. 


95. If we suppose the sphere to be circumscribed by 
a visual cone, the curve in which the perspective plane 
intersects the cone will be the perspective of the sphere. 
The point of sight and the place of the sphere being 
given, the perspective plane may be taken in such a 
position as to intersect the visual cone in any of the 
conic sections. 

Although it is easy to understand why the perspective 
of a sphere may be.an ellipse, a parabola, or an hyper- 
bola; yet neither of these curves seems to be a proper 
representation of a body perfectly round. It is for this 
reason that the perspective of a sphere is generally 
drawn on a plane perpendicular to the line joining the 
point of sight and its centre; when the perspective plane 
has this position, it intersects the visual cone in a circle. 

Let AB (PI. 18), be the ground line, S the centre of 
the picture, D the vanishing point of diagonals, R the 
vanishing point of rays, and H the vanishing point of 
horizontal projections of rays. 

Suppose the centre of the sphere to be at a distance 
behind the perspective plane, equal to its radius. 

Let 5S be the projection of its centre, and the circle 
described with the radius SC the projection of the 
sphere on the perspective plane. 

If through the point of sight we suppose a plane to 
be passed perpendicular to the horizontal and perspec- 
tive planes, LSa' will be its trace on the perspective 

12 


132 TREATISE ON 


plane. This plane will intersect the sphere in a great 
circle, and the visual cone in two elements which will 
be tangent to it. Let this plane be revolved about its 
trace LSa’, until it coincides with the perspective plane. 
The point of sight will fall at D (76), and the centre of 
the circle cut out of the sphere at C. With Cas a 
centre, and radius CS, describe the semicircle, as in the 
figure; and then draw from D the tangent line Da‘a, 
which will be, in its revolved position, an element of the 
visualcone. ‘This element pierces the perspective plane 
at a’; hence, the circle described with the radius Sa’, is 
the perspective of the sphere. 

It is now required to find the perspective of the curve 
of shade. 

It will first be necessary to find the vanishing line of 
the plane of the circle of shade (80). 

Since the plane of shade is perpendicular to the direc- 
tion of the light (34), the required line is the trace of 
a plane passing through the point of sight and perpen- 
dicular to the rays. 

Through the point of sight let a plane of rays be 
passed perpendicular to the perspective plane; RSE is 
its trace. Let this plane be revolved about RE, to coin- 
cide with the perspective plane. ‘The point of sight 
falls at F, in a perpendicular to RS, and at a distance 
from 5 equal to SD. 

Since the ray through the point of sight pierces the 
perspective plane at R, it will, after the revolution, take 
the position RF. If then FE be drawn perpendicular 
to RF, it will be a line of the plane through the point of 
sight, and perpendicular to the direction of the ray, and 
E will be a point of its trace. But RE is the projection, 
on the perspective plane, of the ray passing through the 
point of sight (85): and since the plane through the 


LINEAR PERSPECTIVE. 133 


point of sight is perpendicular to this ray, the trace will 
be perpendicular to its projection. Therefore, GEI, 
drawn perpendicular to RE, is the vanishing line of the 
plane of shade. 

If we suppose two planes of rays to be drawn tangent 
to the visual cone which determines the perspective of 
the sphere, they will also be tangent to the sphere in 
space, and the points of contact will be points of the 
curve of shade. ‘The perspective of these points of 
contact will be found in the perspective of thé sphere, 
and in the traces of the tangent planes. Since the 
tangent planes are planes of rays, they will contain the 
ray through the point of sight; hence their traces will 
pass through the point R. But these traces must also 
be tangent to the perspective of the sphere; hence Ré 
and Re, drawn tangent to the perspective of the sphere, 
are the traces of the tangent planes of rays, and 6 and e 
are two points of the perspective of the circle of shade. 
And since S is the perspective of the centre of the 
sphere, the lines 6Sd and cSe are the indefinite perspec- 
tives of two diameters of the circle of shade. 

Let us suppose, for a moment, the circle of shade to 
‘be represented by the circle dcebg, Fig. n, and let dd and 
ce be the diameters already referred to. The traces of 
the tangent planes of rays are the tangents nd and ne; 
and since the tangent planes of rays are perpendicular 
to the plane of shade, they will intersect in a ray of light 
perpendicular to the plane of the circle poge at n. If 
through this ray and the centre a, a plane be passed, its 
trace ng will bisect the angle cab and be parallel to the 
lines 6 e and cd, joining the corresponding extremities of 
the diameters dd and ce. Hence, the plane of rays 
whose trace is RE, intersects the plane of the circle of 
shade in a line making equal angles with the diameters 


134 TREATISE ON 


whose perspectives are éd and ce. But this line is 
parallel to the chords joining the corresponding ex- 
tremities of these diameters, and also to the line FE, in 
which the plane of rays RE intersects the paralle! plane 
through the point of sight. Therefore, E is the common 
vanishing point of the trace on the plane of shade, and 
of the parallel chords joing the corresponding ex- 
tremities of the diameters. Draw 4E and cK. The 
points e andd in which they intersect cSe and 65d, are the 
perspectives of the extremities of these diameters, and 
consequently, of two more points of the curve of shade. 

To find other points of the curve of shade: 

Let the plane passing through the point of sight and 
parallel to the plane of shade be revolved about its trace 
GI, to coincide with the perspective plane. The point 
of sight falls at F’, a distance from E equal to EF. 
Through this point let any two lines, as F’G and F', be 
drawn at right angles to each other, and note the points 
G and lin which they meet the trace GI. Let us now 
consider the plane to be revolved back to its position in 
space. 

If through the two extremities of any diameter of 
the circle of shade, two lines be drawn parallel to F’G 
and F'l, they will be contained in the plane of shade, 
and their point of intersection on the surface of the 
sphere will be a point of the circle of shade. But these 
two lines will have G and I for their vanishing points (71). 
If therefore, through the points d and 6, we draw the lines 
dG and Ol, they will be the perspectives of two lines 
drawn through the extremities of a diameter at right 
angles to each other, and the point g,in which they 
intersect, is the perspective of a point of the circle of 
shade. The lines through the points ¢ and e determme 


the point f. 


LINEAR PERSPECTIVE 135 


if through /, we draw Sh, it will be the indefinite per- 
spective of a diameter of the circle of shade. Through 
e draw cl. This line is the perspective of the chord 
parallel to the chord whose perspective is fe; therefore, 
his the perspective of the other extremity of the diam- 
eter, and consequently, the perspective of a point of the 
circle of shade. 

Having found a sufficient number of points of the 
curve of shade, let it be described. Only the part cfgé, 
which is in front of the circle of contact of the visual 
cone and sphere, is seen. 

It is now required to find the perspective of the shadow 
cast on the horizontal plane. 

It will first be necessary to find the perspective of the 
horizontal trace of the plane of shade. 

Since this trace is a horizontal line, its vanishing point 
is in the line ‘TSH (74), and since it is a line of the plane 
of shade, its vanishing point is in the line GI; hence it 
is at ‘T’. 

It is necessary in the next place to find the perspective 
of the horizontal projection of the centre of the sphere. 
To do this, lay off from L to P the radius SC of the 
sphere; P is the point at which the diagonal through the 
horizontal projection of the centre of the sphere pierces 
the perspective plane, and the perpendicular through the 
same point pierces it at L; hence zis the perspective 
of the horizontal projection of the centre of the sphere. 

We will here premise, in order to illustrate what fol- 
lows, that when a line intersects the perpendicular from 
the point of sight to the perspective plane, its perspec- 
tive and its projection on the perspective plane are the 
same line: for, the visual plane which determines its 
perspective is then perpendicular to the perspective 
plane. 


136 TREATISE ON 


A second point in the perspective of the horizonta 
trace of the plane of shade is found, by finding the per- 
spective of the point in which any diameter of the circle 
of shade pierces the horizontal plane. ‘To simplify the 
construction, we will take that diameter which is parallel 
to the perspective plane. 

Since the diameter is perpendicular to the ray through 
the centre of the sphere, and since the perspectives of 
the two lines are the same as their projections, it fol- 
lows, that their perspectives will be at right angles to 
each other (Des. Geom. 51). But SR is the perspective 
of the ray; hence, SF drawn perpendicular to SR, is the 
indefinite perspective of the diameter. The projection 
of this diameter on the horizontal plane, passes through 
the horizontal projection of the centre of the sphere, 
and is parallel to the ground line; hence, its perspective 
passes through z and is parallel to AB (72). Therefore, 
k is the perspective of the point in which the diameter 
pierces the horizontal plane, and consequently, a point 
in the perspective of the horizontal trace of the plane 
of shade; and Té is the perspective of that trace. 

The line 7H is the perspective of the horizontal pro- 
jection of the ray through the centre of the sphere, and 
SR is the perspective of the ray; hence p, their point 
of intersection, is the perspective of the shadow cast on 
the horizontal plane by the centre of the sphere. 

The shadow cast on the horizontal plane by any dia- 
meter of the circle of shade, will pass through the point 
in which the diameter pierces the horizontal plane, and 
also through the point of which p is the perspective. 
Therefore, produce the diameter 6d till it meets Tg, the 
perspective of the horizontal trace of the plane of shade, 
and from r draw rd‘pb’; this line is the perspective 
of the indefinite shadow cast by the diameter on the 


LINEAR PERSPECTIVE. 137 


horizontal plane. Through the extremities 6 and d of 
the diameter, draw lines to R; the points 4’ and d’ are 
points of the perspective of the shadow. The diameter 
ec being produced to gq, gives the points ¢ and¢. The 
perspective of the shadow will be tangent to the lines 
Ré and Re at the points 6’ and ¢. 

96. We can find, by a direct construction, the axes of 
the ellipse, which is the perspective of the circle of 
shade. 

That the figure may not become too complicated, we 
will make the construction in Fig. m, in which the pro- 
jection and perspective of the sphere are both repre- 
sented, and in which R is the vanishing point of rays. 

If through the axis of a scalene cone, having a cir- 
cular base, a plane be passed perpendicular to the base, 
it wil divide the cone into two symmetrical parts. If 
then, a plane be passed perpendicular to the plane 
throvgh the axis, it will intersect the cone m a curve 
whose axis is the intersection of the two planes. The 
second axis of the curve 1s the line drawn through the 
middle point of the first, and perpendicular to 1t. 

The visual cone, which is formed by drawing visual 
rays to all the points of the circle of shade, is a scalene 
cone with a circular base. ‘The plane of rays, whose 
trace is SR, passes through the axis, is perpendicu- 
lar to the plane of shade, and also to the perspective 
plane. Hence, the line RS, in which it intersects the 
perspective plane, contains an axis of the ellipse in 
which the perspective plane intersects the visual cone, 
or an axis of the ellipse which is the perspective of the 
circle of shade. ‘The plane, whose trace is RS, also 
intersects the plane of the circle of shade, in a diameter 
perpendicular to the direction ofthe light. Let this plane 
be revolved about RS to coincide with the perspective | 


~ 138 TREATISE ON 

plane. The point of sight falls at F; RF is the revolved 
position of the ray, the centre of the sphere falls at s, and 
psq, drawn perpendicular to RE’, is the diameter of the 
circle of shade. Through p and q draw Fp and Fq; 
Pq is evidently the perspective of pg, and is an axis of 
the ellipse. Through a, the middle point of p‘g, draw 
Faa’. The chord of the circle of shade, whose perspec- 
_ tive is the other axis of the ellipse, passes through the 
point a’, in its true position in space, and is perpendicular 
‘to pag. But since the circle of shade is a great circle 
of the sphere, the length of the chord is equal to cab. 
Let ca'b be revolved about the middle point @’, till it has 
the position cab’, parallel to fR. Through F draw Fe’ 
and I'd’; the points f and d, in which they intersect fR, 
limit the other axis of the ellipse. But this axis must 
_pass through a, and be perpendicular to p’q’. There- 
fore, drawing the perpendicular fd’, and making ad’ and 
af’ equal to af, or ad, we have the second axis of the 
curve. 


PROBLEM X. 


Lo find the perspective of the groined arch and the perspec- 


tive of wuts shadows. 


97. ‘The arch is supported by four pillars, capped by 
cornices, and standing on pedestals; the pedestals are 
placed in the four angles of a square. The arch itself 
is formed by portions of two equal cylinders. 

Fig. n (PI. 19) represents the two projections of the 
pedestals, the pillars, the cornices, and the arch. On the 
line (ab, a’b’), joining the points in which the inner front 
edges of the pillars, produced, pierce the upper plane of 
the cornices, let the semicircle a'sb’ be described, its plane 


Gi 


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LINEAR PERSPECTIVE. 139 


coinciding with that of the front faces of the pillars; 
and on de let there be described an equal and parallel 
semicircle. 

[If now aright line be moved from the position (ad, ’), 
parallel to itself, and touching the two semicircles, it will 
generate a cylinder whose axis is (cm, ¢’), and whose 
elements are perpendicular to the front face of the 
arch. Onthe two lines gf and rv, joining the points in 
which the edges of the pillars pierce the upper plane of 
the cornices, let two equal and parallel semicircles be 
described, and let a right line be moved along them, 
parallel to itself, generating a semi-cylinder whose axis 
zx is parallel to the front face of the arch, or perpendi- 
cular to its side faces. Ml 

The two cylinders which have been generated are 
equal; their axes are at right angles to each other, and 
the surfaces of the cylinders intersect in two equal semi- 
ellipses, called the groins of the arch. 

These groins spring from the upper plane of the cor- 
nices, the one from n to gq, the other from/to ¢. They 
intersect each other in a point of which o is. the hori- 
zontal projection, and which is at a distance above the 
plane of the cornices equal to the radius ¢’s. . 

In the construction of the arch, only a part of each 
cylinder is used. The parts alond and dgote, are formed 
by the cylinder whose axisis em; andthe parts rlogv and 
gnotf, by the cylinder whose axis is zx; the other parts of 
both cylinders are supposed to be removed. 

If we suppose the outer planes of the pillars to be 
produced above the upper plane of the cornices, they 
will form vertical faces of a prism, whose horizontal 
sections are squares. We shall call this prism the solid 
part of the arch. ; 

Let the perspective plane be taken to coincide with 


{40 TREATISE ON 


the front faces of the pedestals and cornices, and let 
AB be the ground line. Let S be the centre of the 
picture, D’ and D the vanishing points of diagonals, R 
the vanishing point of rays, and H the vanishing point 
of horizontal projections. From A and B lay off dis- 
tances equal to a side of the squares which form the 
bases of the pedestals, and on them describe rectangles 
equal to the front faces of the pedestals. ‘Then, let the 
perspectives of the pedestals be found as in Prob. 2. 

From a lay off ab equal to the distance which the 
pedestal projects beyond the pillar. From a draw aD 
to the vanishing point of diagonals, and from 6 draw 6S 
to the vanishing pomt of perpendiculars; d is the per- 
spective of the point in which an edge of the pillar 
pierces the pedestal: and the horizontal line through d is 
the indefinite perspective of the line in which the front 
face of the pillar intersects the pedestal. 

From the other extremity of the line aé lay off a dis- 
tance equal to ab, and draw a Ime to S; the point e, in 
which it intersects de, is the perspective of the point in 
which a second edge of the pillar pierces the pedestal, 
and the point in which it intersects the diagonal aD, is 
the perspective of the point in which a third edge pierces 
the pedestal. ‘Through this latter point draw a parallel 
to de; the point in which it intersects 45 is the perspec- 
tive of the point in which the fourth edge of the pillar 
pierces the pedestal ‘The four vertical lines drawn 
through these points are the indefinite perspectives of the 
vertical edges of the pillar. The front and inner side 
faces only are seen. 

From a draw a vertical line in the perspective plane, 
and on it lay off ac equal to the distance between the 
upper plane of the pedestals and the lower plane of the 
cornices, and through ¢ draw an indefinite horizontal 


LINEAR PERSPECTIVE. 14] 


line cc’. Since the front faces of the cornices are in the 
perspective plane, we can lay off the thickness of the cor- 
nices cc’, and their width cf, equal to the width of the 
pedestals. Project the point 4 into the lower plane of 
the cornices at 6’,and through J draw 8S. The point in 
which this line intersects the edge of the column through 
1,1s the perspective of the point in which that edge 
pierces the lower plane of the cornice; and the hori- 
zontal line through this point is the perspective of the 
line in which the front face of the column intersects the 
lower plane of the cornice. From the point in which 
this horizontal line intersects the edge of the column 
through e, draw a line to S; this line is the perspective 
of the intersection of the inner side face of the column 
with the cornice. The line drawn from ¢ to S is the 
indefinite perspective of the edge of the cornice which 
is perpendicular to the perspective plane at c; the part 
which is seen is limited by the edge of the pillar. 

From f, draw fS, and from c, a diagonal cD; their 
point of intersection /’, is the perspective of the angle 
of the cornice, diagonally opposite to c. Through f 
draw a vertical line, and from the upper extremity of the 
perpendicular through f, draw aline toS; their point of 
intersection is the perspective of an angular point of the 
cornice. ‘The horizontal line through /’ is the perspec- 
tive of the edge parallel to cf’ We have now found the 
perspective of one pillar and one cornice. ‘The perspec- 
tives of the others are found by similar constructions. 

We will next find the perspectives of the front and 
back circles of the arch. 

Find ¢ and e’, the perspectives of the points in which 
the front and inner edges of the pillars pierce the upper 
planes of the cornices; draw ¢e’, and on it describe a 
semicircle, making that part full which is above the 


142 TREATISE ON 


uppet lines of the cornices; this circle is the perspective 
of the front circle of the arch. 

Find the perspectives of the points in which the edges 
of the pillars that are projected at d and e (Fig.n), pierce 
the upper plane of the cornices ; the semicircle described 
on the line joining these points, is the perspective of 
the back circle. 

We next find the perspectives of the groins, and the 
perspectives of the side circles. 

First, find the perspectives of the points », gq, / and ¢ 
(Fig. x), in the upper plane of the cornices, from which 
the groins spring; and also the perspectives of the points 
g, f,r and v, in the same plane, from which the side 
circles spring. 

Let us now suppose the solid part of the arch to be 
intersected by a horizontal plane, and let hi (Fig. ), be 
its trace on the front face of the arch. ‘This plane will 
intersect the solid part of the arch ina square. It will 
intersect the cylinder whose axis is cm, in two elements 
perpendicular to the front face of the arch at the points 
kand p. If through the points in which the horizontal 
projection of either element intersects the lines ng and 
/t, limes be drawn parallel to the ground line, they will 
be the horizontal projections of the elements in which 
the secant plane intersects the other cylinder. The 
points p, p,# and &”, are in the groins, and the square 
k'p'p’k’ has the same diagonals with the square in which 
the secant plane intersects the solid part of the arch. 

Draw any horizontal line, as hz, for the perspective of 
the trace of such a secant plane. From Ah draw AS and 
hD; from 7? draw 7S and7zD’; and from the points & and 
p, draw kS and pS. The points # and p” are the per- 
spectives of the points & and p” (Fig. n), of the groin 


LINEAR PERSPECTIVE. 143 


/t; and the points p’ and &”, the perspectives of the 
points p’ and #” of the groim nq. 

The line fp’ is the perspective of the element /’p’ 
(Fig. x), and therefore the points 4 and 7’, in which it in- 
tersects AS and 7S, are the perspectives of points /’ and 
# (Fig. 7), of the side circles. If we draw a line through 
the points k” and p’, the points h” and 2’, in which it in- 
tersects AS and 2S, are the perspectives of the two points 
h’ and 2” (Fig. n), of the side circles. : 

If we use a second secant plane, we determine two 
other points in the perspective of each groim, and two 
points in the perspective of each of the side circles. 

If through the point of sight a plane be passed tan 
gent to the cylinder whose elements are parallel to the 
perspective plane, it is evident that the perspective of 
the element of contact will pass through the highest 
point of each of the curves. But the perspective of 
this element is a horizontal line (72); hence it will be 
tangent to each of the curves. 

To find this element, and its perspective, suppose a 
vertical plane to be passed through the point of sight, 
and perpendicular to the perspective plane. This plane, 
whose trace is R’g, will intersect the cylinder in a circle 
whose centre is in the upper plane of the cornices, and 
equally distant from the front and back faces of the 
arch. Let this plane be revolved about R’g, to coincide 
with the perspective plane. The point of sight falls at 
D, and the centre of the circle at c’. With c’ asa 
centre, and c’a’, equal to a line corresponding to ca 
(Fig. 2), describe the arc of a circle, and from D draw a 
tangent to this arc. This tangent is a line of the tan- 
gent plane, and the point in which it meets Sg, produced, 
is the point at which it pierces the perspective plane. The 
horizontal line drawn through this point is the perspec- 


144 TREATISE ON 


tive of the element to which the curves are tangent. 
Let the curves be now described. 

The points at which the perspectives of the groins 
intersect the perspectives of the side circles, will limit 
the parts of the side circles which are seen. 

t is now required to find the ‘perspective of the 
shadows cast by the different parts of the arch. 

The front circle of the arch casts a shadow on that 
part of the arch formed by the cylinder whose elements 
are perpendicular to the perspective plane. In finding 
the perspective of this shadow, we have merely to find 
the perspective of the shadow cast by the base of a 
cylinder on the interior surface. 

Throuhg g, draw gl perpendicular to Sik; J 1s the per- 
spective of the pomt at which the shadow on the interior 
of the cylinder begins; for the tangent line at /1is parallel 
to SR, and is the perspective of the trace of a plane of 
rays tangent to the cylinder. ‘To find other points of 
the shadow, draw any line, as ma, parallel to the tangent 
ati. This line is the perspective of the trace of a plane 
of rays on the front face of the arch. This plane inter- 
sects the cylinder in two elements; one casts, and the 
other receives the shadow. ‘The line nS is the perspec- 
tive of the element which receives the shadow, and m’, 
where it is intersected by mR, is the point of shadow. 
Let the curve of shadow from / be then described. 

The rays of light may have such a direction, that the 
shadow of the front circle will intersect the groin k‘p” 
When this occurs, one curve of shadow will pass from 
the point of intersection along the cylinder whose 
elements are parallel to the perspective plane, and will 
pass off on the side circle 2”. Another curve will pass 
from the point of intersection, along the cylinder whose 


LINEAR PERSPECTIVE. 145 


elements are perpendicular to the perspective plane, and 
will pass off on the back circle of the arch. 

The side circle of the arch, on the left, casts a shadow 
on the interior of the arch which falls on the cylinder 
whose elements are parallel to the perspective plane. 
If we suppose a tangent plane of rays to be drawn to 
this cylinder, the point in which the element of contact 
meets the end circle towards the source of light, will be 
the pomt at which the shadow on the interior of the 
cylinder begins. But this tangent plane will be perpen- 
dicular to the side face of the arch; hence, its trace 
will be parallel to the projections of rays on the side 
planes. If we suppose a plane to be passed through R 
parallel to the side planes of the arch, and the ray 
through the point of sight to be projected on this plane, 
the projection will be parallel to the projection of rays 
on side planes, and consequently, to the trace of the 
tangent plane. [f a line be drawn through the point of 
sight parallel to the projection of the ray, it will pierce 
the perspective plane at R’, since RR’ 1s the trace, on the 
perspective plane, of the plane which projects the ray on 
the plane through R; hence R’ is the vanishing point of 
the projections of rays on the side planes. 

If, therefore, through R’ we draw Rq, tangent to the 
curve h’gh’, it will be the perspective of the trace of the 
tangent plane of rays, and gis the perspective of the point 
at which the shadow on the interior of the arch begins. 

if through R’ we draw a secant line, we may regard it 
as the perspective of the trace of a plane of rays, paral- 
lel to the tangent plane, and intersecting the cylinder in 
two elements. 

If through the lower point, in which this line cuts the 
curve A’gh’, we draw a horizontal line, it will be the per- 
spective of the element which receives the shadow; and 


K 


146 TREATISE ON 


drawing through the upper point a line to the vanishing 
point of rays, determines o, the perspective of a point 
of the curve of shadow. Let the curve of shadow be 
then drawn. 

We are next to find the shadows of the cornices on 
the front faces of the pillars. 

Through ¢ draw a line to S, and another to R. The 
point where the line to S meets the line in which the front 
face of the pillar intersects the lower plane of the cor- 
nice, is the perspective of one point in the projection of 
the ray through ¢, on the front face of the pillar, and the 
line drawn through this point, parallel to SR, is the per- 
spective of the projection: the point in which this 
perspective intersects cR is the perspective of one point 
of the line of shadow: but the line of shadow is hori- 
zontal both in space and in perspective. 

To find the shadow cast by the column. The edge er 
casts a line of shadow on the pedestal and on the hori- 
zontal plane, both of which are parallel to the horizontal! 
projection of rays. Hence, their perspectives are easily 
found. At s the shadow falls on the inner side face of 
the back pedestal. It passes up the face in a vertical 
line; reaching the upper face it becomes parallel to the 
horizontal projections of rays, and when it reaches the 
face of the pillar, it ascends in a vertical line along the 
face. If through,, the highest point of the edge of the 
pillar which is in the light, a line be drawn to R, it will 
limit the shadow cast by the edge of the pillar. From 
the point in which this line intersects the vertical line 
before drawn, on the face of the pillar, the shadow will 
be cast by the lower line of the cornice. This shadow 
will be parallel to the projections of rays on the side 
planes, will therefore have its vanishing point at R’, and 
will be limited by the ray through f/ The vertical line 


LINEAR PERSPECTIVE. 147 


through f will then cast its shadow, which will be a 
vertical line, and will be hmited by the ray drawn 
through the upper extremity of the vertical edge of the 
cornice through f. There is a small part of the edge 
of the cornice, parallel to ff’, which is in the light, and 
which will cast a shadow on the face of the pillar. Its 
shadow will be perpendicular to the perspective piane, 
and consequently have its vanishing point at 5. 

The front circle of the arch will next cast a shadow 
on the side face. 

To find this shadow we will observe, that the shadow 
of the diameter ee’, on the side face, is parallel to the 
projections of rays on the side face, and since e” is the 
perspective of the point in which the diameter pierces 
the plane of the side face, e’H’ wii be the indefinite per- 
spective of its shadow. Draw any ordinate of the cir- 
cle, as ¢u, which it is supposed will cast a shadow. 
Through ¢ draw ¢R; the poimt ¢ is the shadow cast on 
the side plane by the foot of the ordinate. But the ordi- 
nate being parallel to the side plane, its shadow will be 
the vertical line through ¢. The point in which this 
vertical line intersects the ray through wis a point of 
shadow on the side plane. When this shadow reaches 
the front face of the pillar, it then falls on that face, and 
is a circle both in space and in perspective. 

Through g draw gR; and through pg’, the perspective 
of the point at which the centre of the front circle is 
projected on the plane of the front faces of the back 
pillars, draw ge” parallel to SR; g” is the perspective 
of the shadow cast by the centre of the front circle 
on the plane of the front faces of’ the back pillars 
But the shadow cast by the radius ge” is parallel to itself; 


“vow 


therefore, draw ge” parallel to the ground line, and e’R 


” wt 


to the vanishing point of rays; g” e” is the perspective 


148 TREATISE ON 


of the radius of the circle of shadow. With g” as a_ 
centre, and g” e” as-a radius, describe the arc of acircle, _ 
and we have the shadow cast on the pillar. The parts 
of the arch which are in the shade, or on which shadows 
fall, are shaded. 


PROBLEM XI. 


It is required to find the perspective of a house ana the perspec- . 
tive of its shadows. 

98. Having measured all the lines of the house which 
are to be put in perspective, make its horizontal pro- 
jection to any convenient scale, as in Pl. 20. 

The outer lines are the horizontal projections of the 
outer lines of the eave-trough, and the adjacent inner 
ones are the lines in which the outer faces of the walls 
intersect the horizontal plane. ‘The lines in which the 
roofs intersect are also made, as well as the projections 
of the chimneys and steps. 

In selecting the position of the perspective plane, 
reference should be had to the part of the building 
which is to appear most prominent in the picture. Were 
it only required to represent the front of the building, 
we should take the perspective plane parallel to it; but 
when the front and an end are to be represented, it is 
most convenient to take the perspective plane oblique to 
them both; though it may be taken parallel to the one 
and perpendicular to the other. It simplifies the con- 
struction, without affecting the generality of the method, 
to assume the perspective plane through one or two of 
the prominent vertical lines. Such lines being in the 
perspective plane will be their own perspectives, and all 
the vertical distances may be laid off on them. 

In the construction here given, the perspective plane 
is passed through the vertical corners of the house 


oo 
v 


™ *? of 


se 


Te ee ee Se ee 


' 
ee 


4a 


he 


+ 3 


i 


: — a foe # 3 


SS 
S=== 


: | x \ \N | ; 
| ee : = ws = hs “4 7 ‘ é- 


hele ee 


il 


| i | ——— 


R 


Ih 


: 


l Ms 


+ 


# 


L\ 
J. 
c »\ 
pn fi Yh | 
x =i = 
lex NS Z a = See ee ie 
e x 4 
: BS TES ey 
ety Bly \ Wx N \. 
aa ts SST AN se \\ 
‘E oie Bi MN = \\' \ e 
| ss -\\ : - a 
— | =< SE \ \ \, =e * . 
oe : os tk . 
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| FS x ne oe = 
as Saye eins | : = 
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= “ \ ~ a < 
. i f 7 pon | \ = 
: | oy mes ; , . 
- ; Fs Soy | | \ 
= es oe Son haf \ 
Y cones | ee Hil \ | \ 
is ae Hl} * ‘ \ y \ 
a i \ 
otee. : ——— eee ies ae Hes ar i = aT < = | : ’ 
7 Ht Sar te, Ripe de \ 
p —_———__,™_!'- ( i \ N Ss Mid a ee l [> \ \ 
—— eae =—— t | i \\ Se iN Ss < 4 = = sah est = 
I | K eet ? | Mi \\ = N < : er 
| 2 ES | j ill \ 771] -: SS | ; | : 
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| i : = ! i \\ OY py cate Seacie \ / conn 
Ae \ ewan S| | A} Neos nay, : “ \ “ é 
= Dich ed | ae ee 77 fT RRS eo eae 
| , | rr £ i | Fe A Wy We : | 2 SS x a N 
: Sa6 | eet ~~ ee i . We IN \ \ \ = 
je H ~] | a \ : a eS Xe x 
ke: | = i /- \ \ \ \ 
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: i Pee / 
| : | | a a - [ 
i Ty J \ % * = ‘ 
| ' es A= \. XX SS \ < 
| “ys \t : ‘ <—T 
| H ! rad Sox ae a 28 a = NX x | ‘ 
1 : ie I! Yt . \ <a \ 
pe a / x x x \ 
| t ie \ S x YE \ \ 
p parse / \ —SSSS ee S JS aries \ 
: fa a \ Be | \ 
i : 2 
; " / / . ae 3 . 
' “ ae 3 es \ N 
ap 
: ¥ ; : 
es 
| 
| 
o 
o| | 
| | 
| 


LINEAR PERSPECTIVE. | oe 6149 


a 


which pierce the horizontal plane at B and A, and DE 
is the ground line. Having chosen the position of the 


eye, lay off the distance FS equal to its height above > 


the ground, and draw KSH, which will be the horizon 
of the picture (74). On the line FG, perpendicular to 
DE, lay off the distance of the point of sight from the 
perspective plane. The horizontal projection of the 
point of sight does not fall on the paper, the eye being 
at about one and a half times the length of the building 
from the perspective plane. In making the drawing, 
a paper must be taken of such dimensions as will enable 

., us to represent upon it the horizontal projection of the 

» point of sight, and the vanishing poate of the principal 
lines. 

Although the horizontal Prema of the house is 
made in front of the ground line, yet the building is in 
fact behind the perspective plane, with its front towards 
the eye. To represent to the mind the horizontal pro- 
jection of the house in its true position, we must con- 
ceive the horizontal plane to be revolved 180° around 
the ground line DE, which will bring every pomt now 
in front of the ground line at an ie distance be- 
hind it. 


At the points a and d draw add Art ded’ perpendi- | 


~ cular to the ground line DE, and make 6 a’ equal to a 8, 
and cd’ equaltocd. The lines Aa and Adjare the lines 
Aa and Ad,in the positions from which their perspectives 
are taken. 

Through the point of sight Jet a line be drawn paral- 
lel to Ad’: its horizontal projection will pass through 
the horizontal projection of the point of sight, and the 
line will pierce the perspective plane 1 in a point of the 
horizontal line KSH. This point is the vanishing point 
of all lines parailel to Ad. The point is not on the 
paper, but as we have frequent occasion to refer to it, 


v 


4 “e 


, 


150 TREATISE ON 


we will designate it by the letter K. Through the point 
of sight let a line be drawn parallel to Aa’: this line 
pierces the perspective plane at H, which is, therefore, 
the vanishing point of all lines parallel to Aa. 

From A draw a line to the vanishing point K; it will 
be the indefinite perspective of the line Ad. From e 
draw a perpendicular to the ground line, and from the 
foot of the perpendicular draw a line to S; the point é, 
in which it intersects the line AK, is one point in the per- 
spective of the corner of the house, which pierces the 
horizontal plane at e; the vertical line drawn through 
“the point ¢ is the indefinite perspective of the corner, 
and the line drawn to B is the perspective of the 
line Be. 

The line drawn from B to the vanishing point K, 1s 
the indefinite perspective of the line Bf From f, draw 
a perpendicular to the ground line, and from the foot of 
the perpendicular draw a line to5; the point’ in which 
it meets the line BK, before drawn, is the perspective of 
the point f ‘Through the point /’, draw a vertical line, 
and it will be the indefinite perspective of the corner of 
the house which pierces the horizontal plane at f/ The 
perspective of the point d is in the line AK, and alsoin 
the line joining ¢c and 8; hence it is atd’ their point of 
intersection. ‘The vertical line drawn through this point 
is the indefinite perspective of the corner of the build- 
ig which meets the horizontal plane at d. 

On the vertical lines passing through the points A and 
B, lay off the distance from the ground .to the eave- 
trough. From the upper extremity of the vertical line 
through A, draw a line to the vanishing point K, and 
.- note the point in which it intersects the vertical line 
through é: the part intercepted between the vertical 
Jines through A and ¢,is the perspective of the lower 


~ “7 


AS {> a 
| Octery $ PAUL 
LINEAR ‘deg sate 151 


line of the eave-trough that is horizontally projected in 
the line Ae. ‘The line joining the upper extremities of 
the vertical lines through b and ¢, is the perspective of 
the lower line of the eave-trough of which Be is the 
projection, and the perspective would, if produced, pass 
through H. From the upper extremity of the vertical 
line through B, draw a line to K. The part cut off by 
the vertical line through /’ is the perspective of the lower 
line of the eave-trough, of which Bf is the horizontal 
projection. ‘Through the upper extremity of the vertical 
line through A, a line has been drawn to K; the part of. 
this line intercepted between the vertical lines through ji 
and d’, is the perspective of all that can be seen a the 
lower line of the eave-trough, whose horizontal projec- 
tion is gd. 

From A draw a line to H, and ane 6 a line to S. 
The vertical line drawn through a’, their point of inter- 
section, 1s the indefinite perspective of the corner of the 
house which meets the horizontal plane at a. From the 
upper extremity of the vertical line through A, draw a 
line to H. The part cut off by the vertical line through 
a’ is the perspective of the lower line of the eave-trough 
of which Aais the horizontal projection. 

On the vertical lines through A and B, lay off the dis- 
tauce to the lower line of the water-table; then the 
width of the water-table; then find the perspectives of 
its upper and lower lines, in the same manner as we have 
already found the perspectives of the lines of the eave- 
trough. On the vertical line through A, lay off from the line © 
DE, the distance to the upper face of the lower window- 
sill, and draw through the point a horizontal line, and 
also a line to K. The horizontal line is the projection ° 
on the perspective plane of the lower line of the windows 
of the first story, and the line to K is its perspective. | 


a» 


Jo2 TREATISE ON 


From the point in which the perspective meets the ver 
tical line through ¢’, draw a line to H, and it will be the 
perspective of the lower line of the window, in that part 
of the building whose horizontal projection is Be: this line 
intersects the vertical line through B, at the same point 
in which itis intersected by the projection of the lower 
line of the windows. ‘The line drawn from this point to 
K is the indefinite perspective ‘of the lower line of the 
windows, corresponding to the part of the house of which 
Bf is the horizontal projection. By similar constructions 
we can find the perspectives of the upper horizontal 
lines of the windows, and also the perspectives of the hori- 
zontal lines which are tangent to their semicircular arches. 

In finding the perspectives of the vertical lines of the 
windows, it should not be forgotten that they are paral- 
lel to the perspective plane, and consequently, their 
perspectives will be parallel to the lines themselves. 
Through the lower extremity of the vertical line whose 
horizontal projection is f, draw a line perpendicular to 
the perspective plane. It will pierce the perspective 
plane in the projection of the lower line of the windows. 
From the point at which it pierces, draw a line to 8: 
its intersection with the perspective of the horizontal 
line before referred to, is the perspective of one point of 
the vertical] bounding line of the window; and the inde- 
finite line can be drawn, since it is vertical. The other 
vertical lines are found by constructions entirely similar, 
and all of them are limited by the perspectives, which 
have’ already been found, of the horizontal lines of the 
building. | 

To find the perspective of the roofs of the house. 
Produce nm, the horizontal projection of the line in 
which the side roofs intersect, until it meets the ground 
line at E. Draw EE’ perpendicular to the ground line, 


+ 


LINEAR PERSPECTIVE. 153 


and make it equal to the height of the intersection of 
the roofs above the ground. Then E’ will be the point 
in which the intersection of the roofs pierces the per- 
spective plane—the horizontal line through E’ is the 
projection, on the perspective plane, of the intersection 
of the roofs, and the line drawn from E’ to the vanish- 
ing point K is its indefinite perspective. ‘The point of 
which £ is the horizontal projection is vertically pro- 
jected atk’. From /' draw alinetoS: the point k’, in 
which it intersects the line E’KX, is the perspective of the 
point (4,£'). ‘The perspectives of the points (m, m’), 
(n,n'), and (p, p’), are found by similar constructions. 

If through the point in which the vertical line through 
A meets the horizontal plane of the eaves of the house, 
a line be drawn to #’, it will be the projection, on the 
perspective piane,of the line in which the side and front 
end roofs, intersect. 

The horizontal lines of the water-table, ofthe windows 
and of the eave-trough, corresponding to the end of the 
house, have a common vanishing point H; and the per- 
spectives of the vertical lines of the windows are found 
in the same manner as those in the front of the house. 

To find the perspectives of the chimneys. [From the 
horizontal line n’E’ lay off a distance above it equal to 
the height of the tops of the chimneys above the inter- 
section of the side roofs, and through the point go de- 
termined, draw the horizontal line xs. Produce the hori- 
zontal lines of the cornices which intersect at cither of 
the angles that are seen, until they meet the perspective | 
plane, which they will do inthe lmers. Since H and K - 
are the vanishing points of these lines, their perspectives 
can be drawn, and the point in which any two of them, 
that pass through the same angular point of the cornice, 
intersect, is the perspective of an angular point of the 


4 


154 TREATISE ON 


cornice. The perspectives of the lines in which the 
faces of the chimneys intersect the roof of the house,and 
the perspectives of their vertical edges, are easily found. 

99. Although the perspective of every point can be 
found rigorously, yet the smaller parts of the building 
can generally be made with sufficient accuracy, without 
making a separate construction for each of them. 
Thus, after we have found the bounding lines of the 
windows, the window-sills, the caps and casings may be 
made very accurately without the aid of a geometrical 
construction. Having found the perspectives of the 
doors and steps, we may make the railing without a par- 
ticular construction for each vertical bar. After having 
found the lower line of the eave-trough, we may draw 
the outer upper line, always observing the rules of per- 
spective, and the general symmetry of the picture. 

In looking obliquely upon a building, the casings of 
the doors and windows on the side nearest the eye are 
not seen, while those on the other side are very distinct. 

To find the perspective of the shadows cast upon 
the house. Let R be the projection of a ray of light 
on the horizontal plane, ft’ its projection on the perspec- 
tive plane, and R” the position of its horizontal projec- 
tion, after it has been revolved to correspond with the 
horizontal projection of the building. ‘The rays of light 
being nearly parallel to the perspective plane, their 
vanishing point will be so far distant from the centre of 
the picture that we cannot use it conveniently in finding 


_the shadows. We therefore adopt other methods. 


In ail the constructions, we must bear in mind that 
the house is behind the perspective plane, and there- 
fore, when a ray of light is drawn through any point, 
its horizontal projection must not be drawn parallel to 
R, but to R’, which makes the same angle with the 


3 . © 
OY CIETY OF oT pA \. 

LINEAR PERSPECTIVE. 159 
ground line, but is differently inclined. Let us first find 
the shadow cast by the upper line of the end eave- 
trough, on the end of the house. 

From ¢ draw a line perpendicular to the ground line, 
and’make vf equal to the distance from the ground to 
the upper line of the eave-trough. Through (¢, #) draw 
aray of light; it pierces the end wall of the house in 
the point (u, wv’), which is a point in the line of shadow. 
From the point in which wu’ meets the ground line, 
draw a line to 8; the point in which it intersects AH is 
the perspective of the pointu. Through this pomt draw 
a vertical line, and it will be the perspective of the vertical 
line passing through (u,v). From uw’, draw a line to 
S; the poimt in which it intersects the vertical line 
before found, is the perspective of one point of the 
required line of shadow. But since the shadow is 
parallel to the line itself, its vanishing point is at H; 
therefore, the line drawn through H, and the point found, 
is the indefinite perspective of the line of shadow. 

Through the highest point, which is in the light, of 
the corner of the house which meets the horizontal 
plane at A, draw a-ray of light. Such ray will pierce 
the face Be of the building in a point which is horizon- 
tally projected at 7. The vertical line drawn through ¢ 
is the shadow cast on the wall by the corner of the 
house which meets the horizontal plane at A. The’ 
perspective of this vertical line of shadow, which is 
easily found, is the perspective of the shadow cast by 
the corner of the house towards the source of light. 
From the upper extremity of this shadow, the shadow 
is cast for a short distance by the upper line of the end 
eave-trough; then by the upper line of the front eave- 
trough; and the shadow terminates in the line of 
shadow cast by the upper line of the eave-trough belong- 


156 TREATISE ON 


ing to the part of the house Be ;.the latter shadow being 
parallel in space to the shadow cast by the end eave- 
trough on the end of the house. 

The shadow of one of the chimneys only is seen. 
This shadow is found by drawing rays of hght thrdugh 
the extremities of the lines casting it; finding the points 
in which they pierce the roof, and determining the per- 
spectives of those points 


GENERAL REMARKS. | 

100. It is, perhaps, too obvious to require illustration, 
that the projections in Descriptive Geometry, viz. the 
Orthographic and Stereographic projections of the 
sphere, are but particular methods of perspective. 

If the point of sight be at an infinite distance from 
the perspective plane, the perspectives of objects are 
the same as their orthographic projections. 

101. If the plane of any circle of the sphere be as- 
sumed for the perspective plane, the point of sight being 


at the pole, the perspective representation of the sphere . 


upon this plane willin nowise differ from its stereographic 
projection. . 

102. We have thus far represented objects on plane 
surfaces only. ‘Their perspectives may however be 
made on other surfaces, and the principles which have 
been explained will require but a slight modification to 
render them applicable to the case in which the per- 
spective is made on any surface whatever. P 

Suppose, for example, it were required to construct 
the perspectives of the different circles of the earth on 


the surface of a tangent cylinder, the point of sight . 


being at the centre.. If we suppose the equator to, be 
the circle of contact, the perspectives of the meridians 


will be elements of the cylinder, and the perspective 


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LINEAR PERSPECTIVE. pT ee 


of any other circle is found by constructing the inter- 
section of the cone, ef which the circle is the base, and 


the point of sight the vertex, with the surface of the . . 
tangent cylinder. Having thus found the perspectives) ._ 


of all the circles of the sphere, if we develop the surface 
of the tangent cylinder on a plane, its development will 
be a map of the earth. Mercator’s chart is constructed 
on these principles. 

103. Panoramic views, which often. exhibit entire 
cities, are generally constructed on the surface of a- 
vertical cylinder, the eye of the spectator being in the 
_ axis. When the perspective is accurately made, and 
viewed from the right point, the deception is perfect. 
The houses seem to stand out from the canvass on 
which they are drawn; the streets have the aspect of 
bustle and business, and one feels himself transported 
into a populous city, and mingling in its affairs. 

104. It is of great importance in linear perspective, 
that the distance of the eye from the plane of the picture 
should be judiciously chosen. If that distance is not 
sufficiently great to enable the eye to command the 
whole surface with perfect convenience, the represen- 
tation, though strictly according to the mathematical 
rules of perspective, will be deceitful. A rule to deter- 
mine this distance is therefore required, and it will be 
found, that-the eye should never approach nearer the 
perspective plane than to a distance equal to the diagonal 
of the picture. 

For example, let CDBA (Pi. 21, Fig. 1), represent 
the picture, then the distance of the point of sight from 
its surface should not be less than AD. I[t may be 
greater, and this is left to the taste of the artist. 

105. Let it be required to put a pavement in perspec- 
tive, composed of alternate squares of black and white 


158 TREATISE ON 


marble—the squares of equal size. Let ABDC (PI. 21, 
Fig. 2), represent the picture; EF its horizon, and G 
the centre of the picture; and let the distance of the 
eye from the perspective plane be equal only to GH. 
Let us now take the same example (PI. 21, Fig. 3), and 
let the distance from the point of sight to the perspec- 
tive plane be equal to the diagonal of the prcture. 

It is evident that, in the first example, the pieces of 
which the pavement is composed neither appear to be 
squares, nor of equal size, while in the second figure 
they do. 

106. The vanishing point of perpendiculars, or the 
point directly opposite the eye; has been called the 
centre of the prcture. ‘This term has been used merely 
to designate the point, and does not imply that the point 
must be situated in the centre of the canvass in which 
the picture is drawn. Indeed, it is not absolutely neces- 
sary that it should be situated within the limits of the 
canvass. It always, however, determines the horizon 
of the picture, which, by a law of nature, rises and falls 
with the eye. 

In the example (PI. 21, Fig. 4), in which AB is the 
horizon, the centre C, of the ae: les without the 
limits of the canvass. 

In the example (PI. 21, Fig. 5), it lies without, and 
below the base line of the picture. 

107. The situation of the horizontal line is altogether 
left to the taste of the artist, and must depend on the 
nature of the subject to be represented. If the appear- 
ance of magnitude in a building, or of majesty in a 
human or ideal figure is required, the choice of a 
very low horizon is the best. It has been remarked 
that the “ Apollo Belvidere was not made to muse in 
a valley.” In a picture, he can only be placed on an 


LINEAR PERSPECTIVE. 159 


eminence, by allowing the horizon to be at his feet or 
below them. 

On the contrary, where a rich back-ground to figures 
is required, it may often be best obtained by placing 
the horizon very high in the picture. In landscape 
painting, the most usual practice is to place the horizon 
rather below than above the middle line of the canvass. 

108. Artists are accustomed to divide linear perspec- 
tive into two kinds, parallel and oblique, and to consider 
the choice of either as entirely a matter of taste. The 
former is the easiest, and therefore from indolence 
would be most frequently chosen. Perhaps, however, 
that in general it looks the best. 


THE END. 


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NOIS-URBANA 


UNIVERSITY OF ILLI 


516.6028111838 c001 
EATISE ON SHADES AND SHAD 


igual 


ows, AND L! 


